332354 + Interview Questions in Secondary School in Math

1.	एक मकान संपत्ति का नगर पालिका मूल्य ₹135000 है तथा उचित किराया 210000 तथा मकान किराया 180000 मकान संपत्ति 20000 प्रतिमाह किराए पर उठाई गई है नगरपालिका हर वर्ष में ₹50000 छिपाये गए हैं मकान गत वर्ष में दो माह खाली है मकान संपत्ति की वार्षिक मूल्य ज्ञात कीजिए
Answer» Answer: NAGAR palika muly is 135000 uchit kiraya is 210000 makan kiraya is 180000 makan sampati is 20000 partemha kiraya is 50000 210000+180000=390000 390000-do mahene khali 390000-50000/12*10 390000-41666 =308334 if you are interested in my answer plz GIVE me LIKE and also follow me then I FOLLOWED you

1.

एक मकान संपत्ति का नगर पालिका मूल्य ₹135000 है तथा उचित किराया 210000 तथा मकान किराया 180000 मकान संपत्ति 20000 प्रतिमाह किराए पर उठाई गई है नगरपालिका हर वर्ष में ₹50000 छिपाये गए हैं मकान गत वर्ष में दो माह खाली है मकान संपत्ति की वार्षिक मूल्य ज्ञात कीजिए

Answer»

Answer:

NAGAR palika muly is 135000

uchit kiraya is 210000

makan kiraya is 180000

makan sampati is 20000

partemha kiraya is 50000

210000+180000=390000

390000-do mahene khali

390000-50000/12*10

390000-41666

=308334

if you are interested in my answer plz GIVE me LIKE and also follow me then I FOLLOWED you

Discussion

2.	Beds are placed for the followingpatients numbered as follows, butthe labels on the beds were flippedaround. The nurses saw these andreported them to the adminstrationoffice.(16,06,68,88,x,98).Whatnumber is the patient X?
Answer» Answer: sorry but I can't UNDERSTAND your QUESTION.. Step-by-step EXPLANATION: PLZZZ follow up

2.

Beds are placed for the followingpatients numbered as follows, butthe labels on the beds were flippedaround. The nurses saw these andreported them to the adminstrationoffice.(16,06,68,88,x,98).Whatnumber is the patient X?

Answer»

Answer:

sorry but I can't UNDERSTAND your QUESTION..

Step-by-step EXPLANATION:

PLZZZ follow up

Discussion

3.	Q5.The range of the data 30, 61, 55,56, 60,20,26, 46,28,56 is..../रिंडेवई 30,61, 55, 56, 60, 20, 26, 46, 28,56 सी हितप्लठमीभा चै।/ दिए गएआंकड़े 30, 61, 55, 56, 60, 20, 26, 46, 28,56 की विचलन सीमाहै ।*O 26O 30O 41O 61
Answer» $Given \: data :$ $30, 61, 55,56, 60,20,26, 46,28,56$ $Maximum \: value = 61$ $Minimum \: value = 20$ $\red{Range}\green { = Maximum \:value - Minimum \: value }$ $= 61 - 20$ $= 41$ THEREFORE., $\red{Range}\green { = 41 }$ •••♪

3.

Q5.The range of the data 30, 61, 55,56, 60,20,26, 46,28,56 is..../रिंडेवई 30,61, 55, 56, 60, 20, 26, 46, 28,56 सी हितप्लठमीभा चै।/ दिए गएआंकड़े 30, 61, 55, 56, 60, 20, 26, 46, 28,56 की विचलन सीमाहै ।*O 26O 30O 41O 61

Answer»

$Given \: data :$

$30, 61, 55,56, 60,20,26, 46,28,56$

$Maximum \: value = 61$

$Minimum \: value = 20$

$\red{Range}\green { = Maximum \:value - Minimum \: value }$

$= 61 - 20$

$= 41$

THEREFORE.,

$\red{Range}\green { = 41 }$

•••♪

Discussion

4.	_______commutative forrational numbers.a) division b) subtraction c) addition d) b and c
Answer» Answer: C) *Addition* Step-by-step explanation: *Google also* *SAYS* so *and* i *also* *KNOW* *this* *only* *about* *this* *question* FUNFACT: FUNFACT:YOU DESERVE HAPPINESS FUNFACT:YOU DESERVE HAPPINESSALWAYS STAY HAPPY PLEASE FIND THE ANSWER IN THE ATTACHED PIC PLEASE REMEMBER TO MARK AS BRAINLIEST IF MY ANSWER IS THE BEST AND RIGHT!**

4.

_______commutative forrational numbers.a) division b) subtraction c) addition d) b and c

Answer»

Answer:

C) Addition

Step-by-step explanation:

Google also SAYS so and i also KNOW this only about this question
FUNFACT:
FUNFACT:YOU DESERVE HAPPINESS
FUNFACT:YOU DESERVE HAPPINESSALWAYS STAY HAPPY
PLEASE FIND THE ANSWER IN THE ATTACHED PIC
PLEASE REMEMBER TO MARK AS BRAINLIEST IF MY ANSWER IS THE BEST AND RIGHT!

Discussion

5.	If the volume of a right circular cone of height 9 cm is 48 t cm', find the diameter of itsbase.A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?The volume of a right circular cone is 9856 cm'. If the diameter of the base is 28 cm,find0 height of the cone(ii) slant height of the cone(ni) curved surface area of the coneAright triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.Find the volume of the solid so obtained.If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then findthe volume of the solid so obtained. Find also the ratio of the volumes of the twosolids obtained in Questions 7 and 8.A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m.Find its volume. The heap is to be covered by canvas to protect it from rain. Find thearea of the canvas required.
Answer» Answer: Q1: height of cone =9cm let radius of cone = R cm given volume of cone = 48 π cm^3 1/3 πr^2h= 48π 1/3 πr^2 *(9) = 48π πr^2 3= 48π r^2 = 48π/3π r^2 = 16 r = √16 r = √(4)² r = 4cm diamemter = 2 ×radius ⇒2×4 = 8cm ∴ the diameter of the cone is 8cm Q2: height of conical pit = H=12m Radius of conical pit = r = diameter ÷ 2 = 3.5÷2 = 1.75m Capacity of pit = volume of cone 1/3 π R² h (1/3 × 22/7 ×1.75 ×12) m³ 38.5 cm³ 38.5 kiloliters.

5.

If the volume of a right circular cone of height 9 cm is 48 t cm', find the diameter of itsbase.A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?The volume of a right circular cone is 9856 cm'. If the diameter of the base is 28 cm,find0 height of the cone(ii) slant height of the cone(ni) curved surface area of the coneAright triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.Find the volume of the solid so obtained.If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then findthe volume of the solid so obtained. Find also the ratio of the volumes of the twosolids obtained in Questions 7 and 8.A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m.Find its volume. The heap is to be covered by canvas to protect it from rain. Find thearea of the canvas required.

Answer»

Answer:

Q1: height of cone =9cm

let radius of cone = R cm

given volume of cone = 48 π cm^3

1/3 πr^2h= 48π

1/3 πr^2 *(9) = 48π

πr^2 3= 48π

r^2 = 48π/3π

r^2 = 16

r = √16

r = √(4)²

r = 4cm

diamemter = 2 ×radius

⇒2×4 = 8cm

∴ the diameter of the cone is 8cm

Q2: height of conical pit = H=12m

Radius of conical pit = r = diameter ÷ 2 = 3.5÷2 = 1.75m

Capacity of pit = volume of cone

1/3 π R² h

(1/3 × 22/7 ×1.75 ×12) m³

38.5 cm³

38.5 kiloliters.

Discussion

6.	Question 2:The two parabolas y = 4ax and y? = 4c(2 - b) cannothave a common normal other than the axis, unless atपैराबोला y' = 4az और y = 40(2-b) अक्ष के अलावा एकसामान्य नहीं हो सकते, जब तक कि
Answer» ANSWER: sjjsgdbxnxnxjxhbxhxhdhdbfbncncjc

6.

Question 2:The two parabolas y = 4ax and y? = 4c(2 - b) cannothave a common normal other than the axis, unless atपैराबोला y' = 4az और y = 40(2-b) अक्ष के अलावा एकसामान्य नहीं हो सकते, जब तक कि

Answer»

ANSWER:

sjjsgdbxnxnxjxhbxhxhdhdbfbncncjc

Discussion

7.	2. 80 హెక్టార్లును 24 రోజులలో 8 మంది కోతకోయగలరు. అయితే36 మంది 30 రోజులలో ఎన్ని హెక్టార్సులో కోతకోయగలరు?
Answer» ANSWER: PLEASE HINDI me LIKHO Step-by-step EXPLANATION:

7.

2. 80 హెక్టార్లును 24 రోజులలో 8 మంది కోతకోయగలరు. అయితే36 మంది 30 రోజులలో ఎన్ని హెక్టార్సులో కోతకోయగలరు?

Answer»

ANSWER:

PLEASE HINDI me LIKHO

Step-by-step EXPLANATION:

Discussion

8.	The length of PB is twice of AB and the length of QC is 3 units more than the lengthof AQ, then let us write, by calculating the length of AC.
Answer» ANSWER: I can't UNDERSTAND the QUESTION

8.

The length of PB is twice of AB and the length of QC is 3 units more than the lengthof AQ, then let us write, by calculating the length of AC.

Answer»

ANSWER:

I can't UNDERSTAND the QUESTION

Discussion

9.	Assignment-6:1. Keeping 6 in the units place and 2 in the crores place, write any five 8-digit numbers usingall the digits 0, 3, 6, 4, 5, 2, 1, 7.(2)(3)(4)(5)13MINARED WORK
Answer» Answer: In place value chart, the digits are grouped in the threes in a big number. The number is read from left to right as ………. BILLION ………. million ……….. THOUSANDS ……….. ones. The place value chart of the International System is given below: 33Save 100,000 = 100 thousand 1,000,000 = 1 million 10,000,000 = 10 millions 100,000,000 = 100 millions Whole numbers can be represented on the place-value chart. 702845 = 7 × 100000 + 2 × 1000 + 8 × 100 + 4 × 10 + 5 × 1 360400295 = 3 × 100000000 + 6 × 10000000 + 4 × 100000 + 2 × 100 + 9 × 10 + 5 × 1 In 702845: Place value of 7 is 7 × 100000 = 700000 and the place is hundred thousand. Place value of 2 is 2 × 1000 = 2000 and the place is thousands. Place value of 8 is 8 × 100 = 800 and the place is hundreds. Place value of 4 is 4 × 10 = 40 and the place is tens. Place value of 5 is 5 × 1 = 5 and the place is ones. In 360400295: Place value of 3 is 3 × 100000000 = 300000000 and the place is hundred million. Place value of 6 is 6 × 10000000 = 60000000 and the place is ten million. Place value of 4 is 4 × 100000 = 400000 and the place is hundred thousand. Place value of 2 is 2 × 100 = 200 and the place is hundred. Place value of 9 is 9 × 10 = 90 and the place is tens. Place value of 5 is 5 × 1 = 5 and the place is ones. Step-by-step explanation: mark as BRAINLIEST answer plzz I need only one more brainliest answer plzz help

9.

Assignment-6:1. Keeping 6 in the units place and 2 in the crores place, write any five 8-digit numbers usingall the digits 0, 3, 6, 4, 5, 2, 1, 7.(2)(3)(4)(5)13MINARED WORK

Answer»

Answer:

In place value chart, the digits are grouped in the threes in a big number. The number is read from left to right as ………. BILLION ………. million ……….. THOUSANDS ……….. ones.

The place value chart of the International System is given below:

33Save

100,000 = 100 thousand

1,000,000 = 1 million

10,000,000 = 10 millions

100,000,000 = 100 millions

Whole numbers can be represented on the place-value chart.

702845 = 7 × 100000 + 2 × 1000 + 8 × 100 + 4 × 10 + 5 × 1

360400295 = 3 × 100000000 + 6 × 10000000 + 4 × 100000 + 2 × 100 + 9 × 10 + 5 × 1

In 702845:

Place value of 7 is 7 × 100000 = 700000 and the place is hundred thousand.

Place value of 2 is 2 × 1000 = 2000 and the place is thousands.

Place value of 8 is 8 × 100 = 800 and the place is hundreds.

Place value of 4 is 4 × 10 = 40 and the place is tens.

Place value of 5 is 5 × 1 = 5 and the place is ones.

In 360400295:

Place value of 3 is 3 × 100000000 = 300000000 and the place is hundred million.

Place value of 6 is 6 × 10000000 = 60000000 and the place is ten million.

Place value of 4 is 4 × 100000 = 400000 and the place is hundred thousand.

Place value of 2 is 2 × 100 = 200 and the place is hundred.

Place value of 9 is 9 × 10 = 90 and the place is tens.

Place value of 5 is 5 × 1 = 5 and the place is ones.

Step-by-step explanation:

mark as BRAINLIEST answer plzz I need only one more brainliest answer plzz help

Discussion

10.	A square park has its side 32 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius 10 m as shown in the figure. Find the area of the remaining part of the park. (Take π = 3.14)
Answer» Here, Required Area = Area of square - Area of 4 quadrants. Radius of ONE Quadrant = 10 m Therefore Area of 4 Quadrants $= \frac{\pi {r}^{2} }{4} \times 4 \\ \\ = 3.14 \times 10 \times 10 \\ \\ = 314 \: \: {m}^{2}$ Now, Required Area $= {32}^{2} - 314 \\ \\ = 1024 - 314 \\ \\ = 710 \: {m}^{2}$ Hence, Area of remaining part = 710 meter SQ.

10.

A square park has its side 32 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius 10 m as shown in the figure. Find the area of the remaining part of the park. (Take π = 3.14)

Answer»

Here, Required Area

= Area of square - Area of 4 quadrants.

Radius of ONE Quadrant = 10 m

Therefore Area of 4 Quadrants

$= \frac{\pi {r}^{2} }{4} \times 4 \\ \\ = 3.14 \times 10 \times 10 \\ \\ = 314 \: \: {m}^{2}$

Now, Required Area

$= {32}^{2} - 314 \\ \\ = 1024 - 314 \\ \\ = 710 \: {m}^{2}$

Hence, Area of remaining part = 710 meter SQ.

Discussion

11.	ਹੇਠਾਂ ਦਿੱਤੇ ਸ਼ਬਦਾਂ/ਮੁਹਾਵਰਿਆਂ ਦੇ ਅਰਥ ਦੱਸ ਕੇ ਵਾਕਾਂ ਵਿੱਚ ਵਰਤੋਂ-(ਉ) ਟੁੰਬਵਾਂ(ਅ) ਹਰਮਨ-ਪਿਆਰਾ(ਏ) ਹੁਲਾਰੇ(ਸ) ਧਮਾਲ(ਹ) ਸਿਖ਼ਰ
Answer» Answer: mujha HINDI or ENGLISH ka alwa or KOI si LANGUAGE samajg mai nhi aati hai Step-by-step explanation: mark me Brainliest please

11.

ਹੇਠਾਂ ਦਿੱਤੇ ਸ਼ਬਦਾਂ/ਮੁਹਾਵਰਿਆਂ ਦੇ ਅਰਥ ਦੱਸ ਕੇ ਵਾਕਾਂ ਵਿੱਚ ਵਰਤੋਂ-(ਉ) ਟੁੰਬਵਾਂ(ਅ) ਹਰਮਨ-ਪਿਆਰਾ(ਏ) ਹੁਲਾਰੇ(ਸ) ਧਮਾਲ(ਹ) ਸਿਖ਼ਰ

Answer»

Answer:

mujha HINDI or ENGLISH ka alwa or KOI si LANGUAGE samajg mai nhi aati hai

Step-by-step explanation:

mark me Brainliest please

Discussion

12.	रिक्त स्थानों की पूर्ति कीजिए-0 (-a) + (6) = (6) + (-0(1) -8 + .................8............. = 0(2)+ [9+(-691 = [2+9] + (........(iv) 15+.= 15
Answer» ANSWER: निम्नलिखित मै परिमेय संख्या है- * 1 point O3 O 2+2/3 O/3/2 0

12.

रिक्त स्थानों की पूर्ति कीजिए-0 (-a) + (6) = (6) + (-0(1) -8 + .................8............. = 0(2)+ [9+(-691 = [2+9] + (........(iv) 15+.= 15

Answer»

ANSWER:

निम्नलिखित मै परिमेय संख्या है-

*

1 point

O3

O

2+2/3

O/3/2

0

Discussion

13.	14) Which is the correct sequence 2 pointsof following activities acoording totime table of your school? *
Answer» ANSWER: ok Step-by-step EXPLANATION:

13.

14) Which is the correct sequence 2 pointsof following activities acoording totime table of your school? *

Answer»

ANSWER:

ok

Step-by-step EXPLANATION:

Discussion

14.	125हेमवालरे7विज्ञानघन टिटिन मैकिता था 3रामसि \|AB B+2P Dan+3C C+5+4+602
Answer» YE KYA LIKHA H?!?!?!?!?!?

14.

125हेमवालरे7विज्ञानघन टिटिन मैकिता था 3रामसि |AB B+2P Dan+3C C+5+4+602

Answer»

YE KYA LIKHA H?!?!?!?!?!?

Discussion

15.	The value of lambda fir which guild u =(X+3y)I +(y-2z)y^+(X+Lambda k^ifp.q=\|p×q\|what is angle between p and q
Answer» ANSWER: gfuyeigufiggfdfggdfgggghh

15.

The value of lambda fir which guild u =(X+3y)I +(y-2z)y^+(X+Lambda k^ifp.q=|p×q|what is angle between p and q

Answer»

ANSWER:

gfuyeigufiggfdfggdfgggghh

Discussion

16.	(a) LCM of two numbers is a factor of their HCF.(b) Product of three numbers is equal to the productof their HCF and LCM.C) HCF of given numbers is always a factor of their LCM.d)LCM of given numbers cannot be smaller than the nune) LCM of co-prime numbers is equal to their product.LCM of two numbers 160 and 352 is 1760. Find their HCF.Write 'True' or 'False' for the following statements.
Answer» ANSWER: true Step-by-step EXPLANATION: because the RESULT of LCM is equal to HCF

16.

(a) LCM of two numbers is a factor of their HCF.(b) Product of three numbers is equal to the productof their HCF and LCM.C) HCF of given numbers is always a factor of their LCM.d)LCM of given numbers cannot be smaller than the nune) LCM of co-prime numbers is equal to their product.LCM of two numbers 160 and 352 is 1760. Find their HCF.Write 'True' or 'False' for the following statements.

Answer»

ANSWER:

true

Step-by-step EXPLANATION:

because the RESULT of LCM is equal to HCF

Discussion

17.	Cost of 5 kg rice is rupees 10 7.501. What will be the cost of 8 kg of rice? 2. What quantity of rice can be purchased in rupees 64.5?
Answer» here's your answer ∴ In RS.107.50 quantity of RICE that can be PURCHASED = Rs. 5 KG ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.50 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.50 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 10750 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100× ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100× 10 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100× 10645 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100× 10645 ∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100× 10645 =3 kg

17.

Cost of 5 kg rice is rupees 10 7.501. What will be the cost of 8 kg of rice? 2. What quantity of rice can be purchased in rupees 64.5?

Answer»

here's your answer

∴ In RS.107.50 quantity of RICE that can be PURCHASED = Rs. 5 KG

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased =

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.50

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased =

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.50

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5=

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 10750

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100×

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100× 10

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100× 10645

∴ In Rs.107.50 quantity of rice that can be purchased = Rs. 5 kg∴ In Rs. 1 quantity of rice that can be purchased = 107.505 kg∴ In Rs. 64.5 quantity of rice that can be purchased = 107.505 ×64.5= 107505 ×100× 10645 =3 kg

Discussion

18.	If (5,k) is a solution of the equation 2x+y-7=0, find the value of kplz give me with the solution so I can understand properly.
Answer» ANSWER: x=5 y=k 2x+y-7=0 2(5)+k=7 10+k=7 k=-10+7 k=-3

18.

If (5,k) is a solution of the equation 2x+y-7=0, find the value of kplz give me with the solution so I can understand properly.

Answer»

ANSWER:

x=5

y=k

2x+y-7=0

2(5)+k=7

10+k=7

k=-10+7

k=-3

Discussion

19.	Factorise:1. 9x?- 16y?1252.3. 81 - 16x?5. 2x-327. 3r-48x9. x-64711. 150 - 6x213. 20x? - 4515. a? - 62 - a -617. a -62 +2bc-?19. a? + 2ab + b2-9c221. (a+b)'-a-b23. x² + 2xy + y2 - a² + 2ab – b?25. a-b-a? + b227. 9-a² + 2ab-b229. 1 + 2ab - (a? +63)31. x² - y2 + 6y-933. 9a+3a-86-64624. 5-20x26. 3ab - 243ab28. 27a? - 486210. 8ab2 - 189712. 2-50r?14. (3a +56) 2-4c?16. 4a2-962-2a-3618. 4a2-4b2 + 4a +120. 108a2 - 3(b-c)?22. x² + y2 - z2 - 2xy24. 25x? - 10x +1-36y?26. a? - 62 - 4ac + 4c?28. x? - 5x? - x +530. 9a? +62 +1 -366232. 4r? -9y? - 2x-3y14. x²+3436. x +4X357-2137. x-139. 81x*-y438. 16r-140. x4-625
Answer» ANSWER: ejieoeoekssjskzsjsjksksksnsjjs

19.

Factorise:1. 9x?- 16y?1252.3. 81 - 16x?5. 2x-327. 3r-48x9. x-64711. 150 - 6x213. 20x? - 4515. a? - 62 - a -617. a -62 +2bc-?19. a? + 2ab + b2-9c221. (a+b)'-a-b23. x² + 2xy + y2 - a² + 2ab – b?25. a-b-a? + b227. 9-a² + 2ab-b229. 1 + 2ab - (a? +63)31. x² - y2 + 6y-933. 9a+3a-86-64624. 5-20x26. 3ab - 243ab28. 27a? - 486210. 8ab2 - 189712. 2-50r?14. (3a +56) 2-4c?16. 4a2-962-2a-3618. 4a2-4b2 + 4a +120. 108a2 - 3(b-c)?22. x² + y2 - z2 - 2xy24. 25x? - 10x +1-36y?26. a? - 62 - 4ac + 4c?28. x? - 5x? - x +530. 9a? +62 +1 -366232. 4r? -9y? - 2x-3y14. x²+3436. x +4X357-2137. x-139. 81x*-y438. 16r-140. x4-625

Answer»

ANSWER:

ejieoeoekssjskzsjsjksksksnsjjs

Discussion

20.	5. Which of the following is not true?(a) (7 + 8) + 9 = 7 + (8 + 9)(b) (7 x 8) x 9 = 7 x (8 x 9)(c) 7 + 8 x 9 = (7 + 8) x (7 + 9)(d) 7 x (8 + 9) = (7 x 8) + (7 x 9)
Answer» Step-by-step explanation: (c) is correct (a) 15+9 = 7+17 24 = 24 (b) 56 ×9 = 7×72 504 = 504 (c) 7 + 72 = 15 × 16 79 ≠ 240 (d) 7 × 17 = 56 + 63 119 = 119 please make me BRAINLIST Dear and follow me

20.

5. Which of the following is not true?(a) (7 + 8) + 9 = 7 + (8 + 9)(b) (7 x 8) x 9 = 7 x (8 x 9)(c) 7 + 8 x 9 = (7 + 8) x (7 + 9)(d) 7 x (8 + 9) = (7 x 8) + (7 x 9)

Answer»

Step-by-step explanation:

(c) is correct

(a) 15+9 = 7+17

24 = 24

(b) 56 ×9 = 7×72

504 = 504

(c) 7 + 72 = 15 × 16

79 ≠ 240

(d) 7 × 17 = 56 + 63

119 = 119

please make me BRAINLIST Dear and follow me

Discussion

21.	In the fiquare area of and is 10cm^2 and the area of triangle adc is 12cm^2a)find BD:DCb) if BD=5cm then find DC
Answer» ANSWER: ok Step-by-step EXPLANATION:

21.

In the fiquare area of and is 10cm^2 and the area of triangle adc is 12cm^2a)find BD:DCb) if BD=5cm then find DC

Answer»

ANSWER:

ok

Step-by-step EXPLANATION:

Discussion

22.	Find the LCM of each of the following group of numbers by prime factorisation.(c) 300, 400, 700
Answer» ANSWER: L.C.M. of 18, 24 and 96 (i) By prime factors Prime factors of 18 = 2 x 3 x 3 Prime factors of 24 = 2 x 2 x 2 x 3 Prime factors of 96 = 2 x 2 x 2 x 2 x 2 x 3 L.C.M. = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288 By common DIVISION method L.C.M. of 18, 24 and 96 = 2 x 2 x 2 x 3 x 3 x 4 = 288 (ii) 100, 150 and 200 Factor of 100 = 2 × 2 × 5 × 5 = 22 × 52 Factor of 150 = 2 × 3 × 5 × 5 = 21 × 31 × 52 Factor of 200 = 2 × 2 × 2 × 5 × 5 = 23 × 52 ∴ L.C.M. OF 100, 150 and 200 = 23 × 31 × 52 = 600 ∴ L.C.M. OF 100, (iii) 14, 21, 98 Factor of 14 = 2 × 7 = 21 × 71 Factor of 21 = 3 × 7 = 31 × 71 Factor of 98 = 2 × 7 × 7 = 21 × 72 ∴ L.C.M. of 14, 21 and 98 = 21 × 31 × 72 = 294 (v) 34, 85 and 51 Factor of 34 = 2 ×17 = 21 × 171 Factor of 85 = 5 × 17 = 51 × 171 Factor of 51 = 3 × 17 = 31 × 171 ∴ L.C.M. of 34, 85 and 51 = 21 × 51 × 31 × 17 = 510 ∴ L.C.M. OF 34, 85 and 51 = 2 × 17 × 5 × 3 = 510

22.

Find the LCM of each of the following group of numbers by prime factorisation.(c) 300, 400, 700

Answer»

ANSWER:

L.C.M. of 18, 24 and 96

(i) By prime factors

Prime factors of 18 = 2 x 3 x 3

Prime factors of 24 = 2 x 2 x 2 x 3

Prime factors of 96 = 2 x 2 x 2 x 2 x 2 x 3

L.C.M. = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288

By common DIVISION method

L.C.M. of 18, 24 and 96 = 2 x 2 x 2 x 3 x 3 x 4 = 288

(ii) 100, 150 and 200

Factor of 100 = 2 × 2 × 5 × 5 = 22 × 52

Factor of 150 = 2 × 3 × 5 × 5

= 21 × 31 × 52

Factor of 200 = 2 × 2 × 2 × 5 × 5 = 23 × 52

∴ L.C.M. OF 100, 150 and 200

= 23 × 31 × 52 = 600

∴ L.C.M. OF 100,

(iii) 14, 21, 98

Factor of 14 = 2 × 7 = 21 × 71

Factor of 21 = 3 × 7 = 31 × 71

Factor of 98 = 2 × 7 × 7 = 21 × 72

∴ L.C.M. of 14, 21 and 98

= 21 × 31 × 72 = 294

(v) 34, 85 and 51

Factor of 34 = 2 ×17 = 21 × 171

Factor of 85 = 5 × 17 = 51 × 171

Factor of 51 = 3 × 17 = 31 × 171

∴ L.C.M. of 34, 85 and 51

= 21 × 51 × 31 × 17 = 510

∴ L.C.M. OF 34, 85 and 51

= 2 × 17 × 5 × 3 = 510

Discussion

23.	There are 28 plants in a garden, of which have thoms in their stems. Howmany plants do not have thorns in theit stems?
Answer» Answer: 28 plants in Garden 2 /7 have thorns in their stems. How MANY do not have thorns? • LET's Head to the QUESTION Now : ⇒ Have Thorns = Total × 2 /7 ⇒ Have Thorns = 28 × 2 /7 ⇒ Have Thorns = 4 × 2 ⇒ Have Thorns = 8 • Plants without thorns in their stem : ⇒ Do Not Have Thorns = Total Plants – Have Thorns ⇒ Do Not Have Thorns = 28 - 8 ⇒ Do Not Have Thorns = 20 ∴ Plants without thorns in stems are 20. _________________________________ • SHORTCUT Trick : ⇒ Do Not Have Thorns = Total × (1 - 2/7) ⇒ Do Not Have Thorns = 28 × (7 - 2) /7 ⇒ Do Not Have Thorns = 28 × 5 /7 ⇒ Do Not Have Thorns = 4 × 5 ⇒ Do Not Have Thorns = 20 ∴ Plants without thorns in stems are 20.

23.

There are 28 plants in a garden, of which have thoms in their stems. Howmany plants do not have thorns in theit stems?

Answer»

Answer:

28 plants in Garden

2 /7 have thorns in their stems.

How MANY do not have thorns?

• LET's Head to the QUESTION Now :

⇒ Have Thorns = Total × 2 /7

⇒ Have Thorns = 28 × 2 /7

⇒ Have Thorns = 4 × 2

⇒ Have Thorns = 8

• Plants without thorns in their stem :

⇒ Do Not Have Thorns = Total Plants – Have Thorns

⇒ Do Not Have Thorns = 28 - 8

⇒ Do Not Have Thorns = 20

∴ Plants without thorns in stems are 20.

_________________________________

• SHORTCUT Trick :

⇒ Do Not Have Thorns = Total × (1 - 2/7)

⇒ Do Not Have Thorns = 28 × (7 - 2) /7

⇒ Do Not Have Thorns = 28 × 5 /7

⇒ Do Not Have Thorns = 4 × 5

⇒ Do Not Have Thorns = 20

∴ Plants without thorns in stems are 20.

Discussion

24.	9.find the areaarea of thepetagon having ventice(1,5), (-2,4) (-3,-1) (-2, -3) (5,1)
Answer» Step-by-step EXPLANATION: SORRY I don't UNDERSTAND your QUESTION ❤❣❣❣❣❣❣❣❣❣

24.

9.find the areaarea of thepetagon having ventice(1,5), (-2,4) (-3,-1) (-2, -3) (5,1)

Answer»

Step-by-step EXPLANATION:

SORRY I don't UNDERSTAND your QUESTION ❤❣❣❣❣❣❣❣❣❣

Discussion

25.	The volume of a cylinder is 69300 cm and its height is 50 cm. Find its diameter thiscylinder is hollow. What would you say about its capacity?
Answer» ANSWER: capacity is NTHG but volume try to find that ok BYE Step-by-step EXPLANATION: HOPE this is help ful pls mark as brainliest

25.

The volume of a cylinder is 69300 cm and its height is 50 cm. Find its diameter thiscylinder is hollow. What would you say about its capacity?

Answer»

ANSWER:

capacity is NTHG but volume try to find that ok BYE

Step-by-step EXPLANATION:

HOPE this is help ful pls mark as brainliest

Discussion

26.	In figure 10.42, T is a point in the exterior of triangle PQR. Show that PQ + QR + PR < 2 (TP +TQ +TR)Solve this ASAP plzzz
Answer» Yes,,,, 2( TP + TQ + TR ) is greater than PQ + QR + PR.... Mark as brainlist.... ✌✌✌✌

26.

In figure 10.42, T is a point in the exterior of triangle PQR. Show that PQ + QR + PR < 2 (TP +TQ +TR)Solve this ASAP plzzz

Answer»

Yes,,,,

2( TP + TQ + TR ) is greater than PQ + QR + PR....

Mark as brainlist....

✌✌✌✌

Discussion

27.	Hey guys plzz answer fast no spamming no wrong answers plzz...1) the equation is 15+3x+y=0 in the same equation ..if x=0 then y=?in the same equation if x=1 then y=?in the same equation if x=2 then y=?in the same equation if x=-1 then y=?
Answer» ANSWER: Here is the answer => Step-by-step EXPLANATION: *EQUATION* =>15+3x+y=0 => *3x+y=-15 => y=15-3x So the* *equation* is y=-15-3x. x=0,then y=-15-3(0) =-15-0 =-15. 2. x=1,then y=-15-3(1) =-15-3 =-18. 3. x=2,then y=-15-3(2) =-15-6 =-21. 4. x=-1,then y=-15-3(-1) =-15+3 =-12. So *the* *values* of *the* *equation* 15+3y+y=0 *are=> x=0,y=-15. x=1,y=-18. x=2,y=-21 x=-1,y=-12. So these* *are* *the* *REQUIRED* *answers.* HOPE IT HELPS.

27.

Hey guys plzz answer fast no spamming no wrong answers plzz...1) the equation is 15+3x+y=0 in the same equation ..if x=0 then y=?in the same equation if x=1 then y=?in the same equation if x=2 then y=?in the same equation if x=-1 then y=?

Answer»

ANSWER:

Here is the answer =>

Step-by-step EXPLANATION:

EQUATION =>15+3x+y=0

=> 3x+y=-15

=> y=15-3x

So the equation is y=-15-3x.

x=0,then y=-15-3(0)

=-15-0

=-15.

2. x=1,then y=-15-3(1)

=-15-3

=-18.

3. x=2,then y=-15-3(2)

=-15-6

=-21.

4. x=-1,then y=-15-3(-1)

=-15+3

=-12.

So the values of the equation 15+3y+y=0 are=>

x=0,y=-15.
x=1,y=-18.
x=2,y=-21
x=-1,y=-12.

So these are the REQUIRED answers.

HOPE IT HELPS.

Discussion

28.	Log (x + 4) - log (3x - 2) = 0, then x =please anyone help me in finding out the value of x
Answer» ANSWER: VALUE of x is 3 . HOPE this will help you out

28.

Log (x + 4) - log (3x - 2) = 0, then x =please anyone help me in finding out the value of x

Answer»

ANSWER:

VALUE of x is 3 . HOPE this will help you out

Discussion

29.	Find the sum of the pair of integers of -9,-6 and -5,+10.No irrelevant answers please otherwise I'll report it
Answer» Step-by-step explanation: (-9)+(-6)= (-9)-6 = (-15) and, (-5)+(+10)=(-5)+10 =(5) now, (-15)+5=(-10) is the answer.. hope the answer was helpful kindly MARK me as a brainliest and follow me GUYS...

29.

Find the sum of the pair of integers of -9,-6 and -5,+10.No irrelevant answers please otherwise I'll report it

Answer»

Step-by-step explanation:

(-9)+(-6)= (-9)-6

= (-15)

and,

(-5)+(+10)=(-5)+10

=(5)

now,

(-15)+5=(-10) is the answer..

hope the answer was helpful

kindly MARK me as a brainliest

and follow me GUYS...

Discussion

30.	A wall is of the shape of a triangle of sides 60m, 100m and140m that has to be painted. The cost of painting is R.S 450 per m2. What is the total cost for painting this wall?
Answer» Answer: it is available on GOOGLE you can CHECK it out please mark as BRAINLIEST if you FOUND this answer HELPFUL

30.

A wall is of the shape of a triangle of sides 60m, 100m and140m that has to be painted. The cost of painting is R.S 450 per m2. What is the total cost for painting this wall?

Answer»

Answer:

it is available on GOOGLE you can CHECK it out please mark as BRAINLIEST if you FOUND this answer HELPFUL

Discussion

31.	Sonaksi8173your answer ... (a)³ - (2√2b³) (a)³ - (√2b)³ we know, a³ - b³ = (a - b)(a² + ab + b²) so, a³ -( √2b)³ = ( a-√2b)(a² + √2²b² + √2ab)
Answer» ANSWER: THANK you so MUCH for HELPING me....

31.

Sonaksi8173your answer ... (a)³ - (2√2b³) (a)³ - (√2b)³ we know, a³ - b³ = (a - b)(a² + ab + b²) so, a³ -( √2b)³ = ( a-√2b)(a² + √2²b² + √2ab)

Answer»

ANSWER:

THANK you so MUCH for HELPING me....

Discussion

32.	Medicine is packed in boxes.cach weighing 4 kg soog-hanz-manSuch bores can be loaded in a can which cannot consy beyond800 kgorionne sono=4500
Answer» ANSWER:Medicines are packed in boxes. One box weighs 4KG 500g. We need to find many the number of boxes that can be loaded in a can that cannot carry beyond 800kg. Solution: The weight of one box = 4 kg 500 G = 4 × 1000 g + 500 g = 4500 g { As we know that 1kg = 1000G} Maximum load can be loaded in can = 800 kg = 800 × 1000 g = 800000 g Hence total weight=Maximum load can be loaded in can =800000 g Number of boxes = Total weight / Weight of each box Number of boxes = 800000/4500 = 177 Number of boxes = 177 Therefore, 177 boxes can be loaded in a can that cannot carry beyond 800kg.

32.

Medicine is packed in boxes.cach weighing 4 kg soog-hanz-manSuch bores can be loaded in a can which cannot consy beyond800 kgorionne sono=4500

Answer»

ANSWER:Medicines are packed in boxes. One box weighs 4KG 500g.

We need to find many the number of boxes that can be loaded in a can that cannot carry beyond 800kg.

Solution:

The weight of one box = 4 kg 500 G = 4 × 1000 g + 500 g = 4500 g { As we know that 1kg = 1000G}

Maximum load can be loaded in can = 800 kg = 800 × 1000 g = 800000 g

Hence total weight=Maximum load can be loaded in can =800000 g

Number of boxes = Total weight / Weight of each box

Number of boxes = 800000/4500 = 177

Number of boxes = 177

Therefore, 177 boxes can be loaded in a can that cannot carry beyond 800kg.

Discussion

33.	2 5:3 11X.OReciprocal of 11/5 [11/5 रा ठप्लटभ ][11/5 का गुणात्मक प्रतिलोम ]Reciprocal of 5/11 [5/11 रा हप्लटवभ][5/11 का गुणात्मक प्रतिलोम]O - 11/5O - 5/117:57pm
Answer» Question :- Find the RECIPROCAL of 11/5 . Answer :- As we KNOW that, The reciprocal of a number is equal to 1 DIVIDED by the number. So, → reciprocal of 11/5 = 1 ÷ (11/5) → reciprocal of 11/5 = 1 * (5/11) → reciprocal of 11/5 = (5/11) (Ans.) Or, we can say that, when we MULTIPLY a number by its reciprocal we get 1.. So, → (11/5) * reciprocal = 1 → reciprocal = 1/(11/5) → reciprocal = 1 * (5/11) → reciprocal = (5/11) (Ans.)

33.

2 5:3 11X.OReciprocal of 11/5 [11/5 रा ठप्लटभ ][11/5 का गुणात्मक प्रतिलोम ]Reciprocal of 5/11 [5/11 रा हप्लटवभ][5/11 का गुणात्मक प्रतिलोम]O - 11/5O - 5/117:57pm

Answer»

Question :- Find the RECIPROCAL of 11/5 .

Answer :-

As we KNOW that, The reciprocal of a number is equal to 1 DIVIDED by the number.

So,

→ reciprocal of 11/5 = 1 ÷ (11/5)

→ reciprocal of 11/5 = 1 * (5/11)

→ reciprocal of 11/5 = (5/11) (Ans.)

Or,

we can say that, when we MULTIPLY a number by its reciprocal we get 1..

So,

→ (11/5) * reciprocal = 1

→ reciprocal = 1/(11/5)

→ reciprocal = 1 * (5/11)

→ reciprocal = (5/11) (Ans.)

Discussion

34.	If a solid figure has 15 edges and 10 vertices, then the number of faces of a solid figure is(1) 9(2) 7(3) 11(4) 5
Answer» Answer: 11 Step-by-step EXPLANATION: jabsudvduehhe in d VU dirvueof

34.

If a solid figure has 15 edges and 10 vertices, then the number of faces of a solid figure is(1) 9(2) 7(3) 11(4) 5

Answer»

Answer:

11

Step-by-step EXPLANATION:

jabsudvduehhe in d VU dirvueof

Discussion

35.	Factorise 216a³-2root2b³amd please don't write "I don't know"If you know then only answerplease
Answer» $Given \: 216a^{3} -2\sqrt{2} b^{3}$ $= 6^{3} a^{3} - (\sqrt{2})^{3} b^{3}$ $=( 6a)^{3} - (\sqrt{2}b)^{3}$ /* By ALGEBRAIC IDENTITY */ $\boxed{ \pink { x^{3} - y^{3} = (x-y)(x^{2}+xy+y^{2})}}$ $= (6a-\sqrt{2}b)[ (6a)^{2} + (6a)(\sqrt{2}b) + (\sqrt{2}b)^{2} ]$ $= (6a-\sqrt{2}b)(36a^{2} + 6\sqrt{2}ab + 2b^{2} )$ THEREFORE., $\red{ 216a^{3} -2\sqrt{2} b^{3} }$ $\green {= (6a-\sqrt{2}b)(36a^{2} + 6\sqrt{2}ab + 2b^{2} ) }$ •••♪

35.

Factorise 216a³-2root2b³amd please don't write "I don't know"If you know then only answerplease

Answer»

$Given \: 216a^{3} -2\sqrt{2} b^{3}$

$= 6^{3} a^{3} - (\sqrt{2})^{3} b^{3}$

$=( 6a)^{3} - (\sqrt{2}b)^{3}$

/* By ALGEBRAIC IDENTITY */

$\boxed{ \pink { x^{3} - y^{3} = (x-y)(x^{2}+xy+y^{2})}}$

$= (6a-\sqrt{2}b)[ (6a)^{2} + (6a)(\sqrt{2}b) + (\sqrt{2}b)^{2} ]$

$= (6a-\sqrt{2}b)(36a^{2} + 6\sqrt{2}ab + 2b^{2} )$

THEREFORE.,

$\red{ 216a^{3} -2\sqrt{2} b^{3} }$

$\green {= (6a-\sqrt{2}b)(36a^{2} + 6\sqrt{2}ab + 2b^{2} ) }$

•••♪

Discussion

36.	Which of the following sequences are AP? In case of AP, find the first terma and the common difference d and write the next three terms.(i) 1²,2²,3²,4²,5²,...
Answer» Step-by-step EXPLANATION: GIVEN SEQUENCE is not an AP

36.

Which of the following sequences are AP? In case of AP, find the first terma and the common difference d and write the next three terms.(i) 1²,2²,3²,4²,5²,...

Answer»

Step-by-step EXPLANATION:

GIVEN SEQUENCE is not an AP

Discussion

37.	ਦਿੱਠੇ ਅੰਡੋਂ 36 , 6), 55, 56 , 60 , 00 , 26 , 16 . 28 , 5 6ਦੀਵਿਚਲਸੰਸ ਹੈ258.304}
Answer» ANSWER: PLEASE EXPLAIN again...............

37.

ਦਿੱਠੇ ਅੰਡੋਂ 36 , 6), 55, 56 , 60 , 00 , 26 , 16 . 28 , 5 6ਦੀਵਿਚਲਸੰਸ ਹੈ258.304}

Answer»

ANSWER:

PLEASE EXPLAIN again...............

Discussion

38.	IFyears ago my age was 5 times thesonShire.age of my sonthen six years ago, my agewas numerically ture the square of theage of my son then. Find the presentage of myson
Answer» nooooooooooooooooooooooo

Discussion

39.	6. One of the diagonals of a rhombus is equal to one of its sides. Find the angles of the rhombus.7. ABCD is a rhombus. If ZACD = 40°, find ZADB.In a square PQRS, PQ = (5a – 17) cm and QR = (2a + 4) cm. Find the lengths of PS and PR.
Answer» Answer: 90 ° Step-by-step EXPLANATION: Because , the all sides of ROMBUS is congruent so all Angle is congruent MEANS all Angle are 90°

39.

6. One of the diagonals of a rhombus is equal to one of its sides. Find the angles of the rhombus.7. ABCD is a rhombus. If ZACD = 40°, find ZADB.In a square PQRS, PQ = (5a – 17) cm and QR = (2a + 4) cm. Find the lengths of PS and PR.

Answer»

Answer:

90 °

Step-by-step EXPLANATION:

Because , the all sides of ROMBUS is congruent

so all Angle is congruent MEANS all Angle are 90°

Discussion

40.	DOMS Page No.1L.S-4 Angls & pairs AnglesQ. 17 Observe the figure and completethe table for I answers.1.Name of the points in the interior
Answer» Step-by-step EXPLANATION: DIVIDE MULTIPLE SUBTRACT

40.

DOMS Page No.1L.S-4 Angls & pairs AnglesQ. 17 Observe the figure and completethe table for I answers.1.Name of the points in the interior

Answer»

Step-by-step EXPLANATION:

DIVIDE MULTIPLE SUBTRACT

Discussion

41.	Find the vertex of the quadratic function: f(x)=x^2+10x+14The equation for the vertex is:X=-b/2ay=f(-b/2a)Pre-calc Answer!
Answer» ANSWER: ( - 5, - 11) Step-by-step explanation: f(X) = x² + 10x + 14 The coordinates for the vertex are: y = f( - 5) =( - 5)² + 10 × ( - 5) + 14 = - 11 ( - 5, - 11) Zeroes of given FUNCTION are = - 5 ± √11

41.

Find the vertex of the quadratic function: f(x)=x^2+10x+14The equation for the vertex is:X=-b/2ay=f(-b/2a)Pre-calc Answer!

Answer»

ANSWER:

( - 5, - 11)

Step-by-step explanation:

f(X) = x² + 10x + 14

The coordinates for the vertex are:

y = f( - 5) =( - 5)² + 10 × ( - 5) + 14 = - 11

( - 5, - 11)

Zeroes of given FUNCTION are = - 5 ± √11

Discussion

42.	6EXERircus artist is climbing a 20 m long ropched and tied from the top of a verticalthe height of the pole, if the angle macound level is 30° (see Fig. 9.18).breaks due to storm and the broken pthe tree touches the ground making
Answer» Step-by-step explanation: LET AB be the vertical pole and CA be the rope. Then, ∠ACB=30 degreeand AC=20 m In right △ ABC, sin30 DEGREE = AB /AC 1/2= AB/20 AB=10 m Therefore, the height of the pole is 10 m.

42.

6EXERircus artist is climbing a 20 m long ropched and tied from the top of a verticalthe height of the pole, if the angle macound level is 30° (see Fig. 9.18).breaks due to storm and the broken pthe tree touches the ground making

Answer»

Step-by-step explanation:

LET AB be the vertical pole and CA be the rope. Then,

∠ACB=30 degreeand AC=20 m

In right △ ABC,

sin30 DEGREE = AB /AC

1/2= AB/20

AB=10 m

Therefore, the height of the pole is 10 m.

Discussion

43.	From the following data. Find range and quartile deviation. also determine their coefficient.Months 1 2 3 4 5 6 7 8 9 10 11Sales(र) 78 80 80 82 84 84 86 86 88 88 90please give solutions with step by step
Answer» Answer:Here, Highest value (H)= 35 LOWEST value (L) = 15 Range= Highest value − Lowest value i.e. R= H− L Substituting the given values in the formula R= 35 − 15= 20 COEFFICIENT of Range is as follows: CR = H-L H+L or, CR = 35-15 35+15 = 20 50 ⇒ CR =0.4 Hence, the range (R) of the above data is 20 and coefficient of Range (CR) is 0.4

43.

From the following data. Find range and quartile deviation. also determine their coefficient.Months 1 2 3 4 5 6 7 8 9 10 11Sales(र) 78 80 80 82 84 84 86 86 88 88 90please give solutions with step by step

Answer»

Answer:Here,

Highest value (H)= 35

LOWEST value (L) = 15

Range= Highest value − Lowest value

i.e. R= H− L

Substituting the given values in the formula

R= 35 − 15= 20

COEFFICIENT of Range is as follows:

CR =

H-L

H+L

or, CR =

35-15

35+15

=

20

50

⇒ CR =0.4

Hence, the range (R) of the above data is 20 and coefficient of Range (CR) is 0.4

Discussion

44.	SECTION D3V2V15 +3/735Simplify:7V32V5V10 + V3 V6 + v5
Answer» so,it can be 5/12 PLEASE MARK me as BRAILIST

Discussion

45.	The breadth of a rectangle is 5 less than twice its length. If the perimeter of the rectangle is 14 m. findthe dimensions of the rectangle.
Answer» Step-by-step explanation: LET, length of rectangle = X breadth of rectangle = 2x - 5 perimeter of rectangle = 14 m. 2 ( l + b ) = 14 2x - 5 + x = 7 3x = 12 x = 4. 2x - 5 = 2(4) - 5 = 8 - 5 = 3 therefore, length of rectangle = x = 4m. breadth of rectangle = 2x - 5 = 3m.

45.

The breadth of a rectangle is 5 less than twice its length. If the perimeter of the rectangle is 14 m. findthe dimensions of the rectangle.

Answer»

Step-by-step explanation:

LET, length of rectangle = X

breadth of rectangle = 2x - 5

perimeter of rectangle = 14 m.

2 ( l + b ) = 14

2x - 5 + x = 7

3x = 12

x = 4.

2x - 5 = 2(4) - 5 = 8 - 5 = 3

therefore, length of rectangle = x = 4m.

breadth of rectangle = 2x - 5 = 3m.

Discussion

46.	In a class of 175 students the following data shows the number of students opting one or moresubjects. Mathematics 100, Physics 70, Chemistry 40, Mathematics and Physics 30, Mathematics and Chemistry 28, Physics and Chemistry 23, Mathematics, Physics and Chemistry 18. (i) How many students have taken at least one of the three subjects? (ii) How many students have taken exactly one subject? (iii) How many students have taken none of these subjects?
Answer» ANSWER: ............ Step-by-step EXPLANATION: .............

46.

In a class of 175 students the following data shows the number of students opting one or moresubjects. Mathematics 100, Physics 70, Chemistry 40, Mathematics and Physics 30, Mathematics and Chemistry 28, Physics and Chemistry 23, Mathematics, Physics and Chemistry 18. (i) How many students have taken at least one of the three subjects? (ii) How many students have taken exactly one subject? (iii) How many students have taken none of these subjects?

Answer»

ANSWER:

............

Step-by-step EXPLANATION:

.............

Discussion

47.	3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be 160. After amonth, the cost of 4 kg of apples and 2 kg of grapes is 7 300. Represent the situationalgebraically and geometrically.
Answer» ANSWER: I am SURE that the answer is CORRECT. ADD the answer PROPLE. add the answer prople. add the answer prople. add the answer prople. add the answer prople.

47.

3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be 160. After amonth, the cost of 4 kg of apples and 2 kg of grapes is 7 300. Represent the situationalgebraically and geometrically.

Answer»

ANSWER:

I am SURE that the answer is CORRECT. ADD the answer PROPLE. add the answer prople. add the answer prople. add the answer prople. add the answer prople.

Discussion

48.	In how meny ways can three men and three women sit at a round table so that no two men can occupy adjacent positions ?
Answer» Let the 1st WOMAN choose a seat. She has 6 choices and PICKS 1. It becomes the anchor seat. 6 choices then collapse to just 1 because all seating permutations relative to any of the 6 choices available to become the anchor seat are identical relative to the circular seating arrangement.The 2nd woman has a choice of 2 SEATS and the 3rd woman has just 1 choice of seating. The men can then be accommodated with the 3 remaining seats in 3! ways. In response to the QUESTION, the TOTAL of seating arrangements with the constraint as specified = 12(3!) = 12. please mark as the Brainliest answer

48.

In how meny ways can three men and three women sit at a round table so that no two men can occupy adjacent positions ?

Answer»

Let the 1st WOMAN choose a seat. She has 6 choices and PICKS 1. It becomes the anchor seat. 6 choices then collapse to just 1 because all seating permutations relative to any of the 6 choices available to become the anchor seat are identical relative to the circular seating arrangement.The 2nd woman has a choice of 2 SEATS and the 3rd woman has just 1 choice of seating. The men can then be accommodated with the 3 remaining seats in 3! ways.

In response to the QUESTION, the TOTAL of seating arrangements with the constraint as specified = 1*2*(3!) = 12.

please mark as the Brainliest answer

Discussion

49.	If p(x) = ax2 + bx + c, and a + b + c = 0, then one zero(a)-b/a(b) c/a(d) b/c(e)none of these
Answer» ANSWER: *OPTION* a *and* b *both* Step-by-step explanation: ax^2+bx+c=0(A=a, B=b and C=c) ACCORDING to QUADRATIC formula alfa+beta=-b/a alfa×beta=c/a

49.

If p(x) = ax2 + bx + c, and a + b + c = 0, then one zero(a)-b/a(b) c/a(d) b/c(e)none of these

Answer»

ANSWER:

OPTION a and b both

Step-by-step explanation:

ax^2+bx+c=0(A=a, B=b and C=c)
ACCORDING to QUADRATIC formula
alfa+beta=-b/a
alfa×beta=c/a

Discussion

50.	F P() denotes the probability of an event A, thena) P() < 0 b) P() > 1 c) ≤ () ≤ d) − ≤ () ≤
Answer» ANSWER: -1 < P ≤ 1 Step-by-step explanation: hope this will be HELPFUL for you

50.

F P() denotes the probability of an event A, thena) P() < 0 b) P() > 1 c) ≤ () ≤ d) − ≤ () ≤

Answer»

ANSWER:

-1 < P ≤ 1

Step-by-step explanation:

hope this will be HELPFUL for you

Discussion

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