332354 + Interview Questions in Secondary School in Math

1.	(25)^1/2×5^1/3 calculate it
Answer» Step-by-step EXPLANATION: This is REQUIRED SOLUTION to the QUESTION.

1.

(25)^1/2×5^1/3 calculate it

Answer»

Step-by-step EXPLANATION:

This is REQUIRED SOLUTION to the QUESTION.

Discussion

2.	Factorise the following
Answer» ANSWER: $\frac{225x ^{2} - 4y {?}^{2} }{36}$ Step-by-step EXPLANATION: $\frac{25x {?}^{2} }{4} - \frac{y {}^{2} }{9} \\ lcm \: of \: 4 \: and \: 9 = 36 \\ \frac{225x {}^{2} - 4y {}^{2} }{36} answer$ HOPE it HELPS you

2.

Factorise the following

Answer»

ANSWER:

$\frac{225x ^{2} - 4y {?}^{2} }{36}$

Step-by-step EXPLANATION:

$\frac{25x {?}^{2} }{4} - \frac{y {}^{2} }{9} \\ lcm \: of \: 4 \: and \: 9 = 36 \\ \frac{225x {}^{2} - 4y {}^{2} }{36} answer$

HOPE it HELPS you

Discussion

3.	1) In a right angle triangle, the difference between areas of squares drawn onperpendicular sides of triangles is 7 sq.cm., and the area of square drawn onhypotenuse is 25 sq.cm. Find sides of triangle.
Answer» Correct QUESTION :- In a right angle triangle, the areas of square drawn on perpendicular sides of a triangle is 49 Sq. cm. and the areas of square drawn on HYPOTENUSE is 625 Sq. cm. find the sides of the triangle ? SOLUTION :- we know that, side of square = √(Area) so, → Area of square drawn on perpendicular side = 49 cm² → Perpendicular side = √(49) = 7 cm. SIMILARLY, → Area of square drawn on hypotenuse side = 625 cm² → Perpendicular side = √625 = 25 cm. then, → Base = √[(Hypotenuse)² - (Perpendicular)²] → Base = √[(25)² - (7)²] → Base = √(625 - 49) → Base = √576 → Base = 24 cm. therefore, the sides of the triangle are 7 cm , 24 cm and 25 cm.

3.

1) In a right angle triangle, the difference between areas of squares drawn onperpendicular sides of triangles is 7 sq.cm., and the area of square drawn onhypotenuse is 25 sq.cm. Find sides of triangle.

Answer»

Correct QUESTION :- In a right angle triangle, the areas of square drawn on perpendicular sides of a triangle is 49 Sq. cm. and the areas of square drawn on HYPOTENUSE is 625 Sq. cm. find the sides of the triangle ?

SOLUTION :-

we know that,

side of square = √(Area)

so,

→ Area of square drawn on perpendicular side = 49 cm²

→ Perpendicular side = √(49) = 7 cm.

SIMILARLY,

→ Area of square drawn on hypotenuse side = 625 cm²

→ Perpendicular side = √625 = 25 cm.

then,

→ Base = √[(Hypotenuse)² - (Perpendicular)²]

→ Base = √[(25)² - (7)²]

→ Base = √(625 - 49)

→ Base = √576

→ Base = 24 cm.

therefore, the sides of the triangle are 7 cm , 24 cm and 25 cm.

Discussion

4.	110. A metallic cylinder of diameter 5 cm. and height 3 cm. is melted and cast into a sphere.What is its diameter.3
Answer» Answer: GIVEN diameter of metallic cylinder=5 cm So radius of cylinder=2.5 cm And its HEIGHT =3 3 1 = 3 10 cm Then VOLUME of cylinder = πr 2 h=π×(2.5) 2 × 3 10 cm 3 Let radius of sphere is r Then volume of sphere = 3 4 πr 3 Then volume of cylinder =volume of sphere ∴ 3 4 πr 3 =π×2.5×2.5× 3 10 ⇒r 3 = 4 2.5×2.5×10 = 4 62.5 =15.625 ⇒r=2.5cm

4.

110. A metallic cylinder of diameter 5 cm. and height 3 cm. is melted and cast into a sphere.What is its diameter.3

Answer»

Answer:

GIVEN diameter of metallic cylinder=5 cm

So radius of cylinder=2.5 cm

And its HEIGHT =3

3

1

=

3

10

cm

Then VOLUME of cylinder = πr

2

h=π×(2.5)

2

×

3

10

cm

3

Let radius of sphere is r

Then volume of sphere =

3

4

πr

3

Then volume of cylinder =volume of sphere

∴

3

4

πr

3

=π×2.5×2.5×

3

10

⇒r

3

=

4

2.5×2.5×10

=

4

62.5

=15.625

⇒r=2.5cm

Discussion

5.	Define trial and error method.give examples of them .solve the following with error method and trial method.a)7x-5=9b)10x+15=55c)4p by 5 =16..
Answer» Answer: TRIAL and error is a fundamental method of problem-solving. It is characterized by repeated, varied ATTEMPTS which are continued until SUCCESS, or until the practicer stops TRYING.

5.

Define trial and error method.give examples of them .solve the following with error method and trial method.a)7x-5=9b)10x+15=55c)4p by 5 =16..

Answer»

Answer:

TRIAL and error is a fundamental method of problem-solving. It is characterized by repeated, varied ATTEMPTS which are continued until SUCCESS, or until the practicer stops TRYING.

Discussion

6.	The radio of two sphere are 3.5 find the ratio of there surface area and volumes
Answer» ANSWER: I am sorryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6.

The radio of two sphere are 3.5 find the ratio of there surface area and volumes

Answer»

ANSWER:

I am sorryyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Discussion

7.	Triangle ABC is an equilateral triangle, whose side is 2a. Find its height?
Answer» 2a Step-by-step EXPLANATION: Because it is an EQUILATERAL TRIANGLE

7.

Triangle ABC is an equilateral triangle, whose side is 2a. Find its height?

Answer»

2a

Step-by-step EXPLANATION:

Because it is an EQUILATERAL TRIANGLE

Discussion

8.	To find out the concentration of So, in the air (in parts per million, i.e., ppm), the datawas collected for 30 localities in a certain city and is presented below:ncentration of So, in ppm0.00-0.04 0.04-0.08 0.08-0.12 0.12-0.16 0.16-0.20 0.20-0.2424frequency9492.Find the mean concentration of So4, in the air.
Answer» I'm AYUSH uppadhayay 6A and I will do this GAME

Discussion

9.	18-3y/4 =11+y in the book it is written anwer is 4 I what calculation
Answer» Step-by-step explanation: 18-3y/4=11+y 18-3y=4(11+y) 18-3y=44+4y -3y-4y=44-1 -7y=26 y=26/-7

9.

18-3y/4 =11+y in the book it is written anwer is 4 I what calculation

Answer»

Step-by-step explanation:

18-3y/4=11+y

18-3y=4(11+y)

18-3y=44+4y

-3y-4y=44-1

-7y=26

y=26/-7

Discussion

10.	Subtract: 3y - 2xy --- x' fromy - 2 x from 7 y + xy + 2x
Answer» Step-by-step EXPLANATION: DJ CURLS CIVIC crump HARBOR

10.

Subtract: 3y - 2xy --- x' fromy - 2 x from 7 y + xy + 2x

Answer»

Step-by-step EXPLANATION:

DJ CURLS CIVIC crump HARBOR

Discussion

11.	Or-VI, Dwarka, Nei5082166, 2508525nail : dav.dwarkaVebsite : www.da
Answer» Step-by-step EXPLANATION: come rgq-dwag-dch liploplop

11.

Or-VI, Dwarka, Nei5082166, 2508525nail : dav.dwarkaVebsite : www.da

Answer»

Step-by-step EXPLANATION:

come

rgq-dwag-dch

liploplop

Discussion

12.	अधिकारों को कानूनी मान्यता कैसे मिलती हैं
Answer» कि बहुत से सिद्धांतकार अधिकार को ऐसे दावों के रूप में परिभाषित करते हैं, जिन्हें राज्य ने मान्य किया हो। कानूनी मान्यता से यकीनन हमारे अधिकारों को समाज में एक खास दर्जा मिलता है, लेकिन कानूनी मान्यता के आधार पर किसी अधिकार का दावा नहीं किया जा सकता।

12.

अधिकारों को कानूनी मान्यता कैसे मिलती हैं

Answer»

कि बहुत से सिद्धांतकार अधिकार को ऐसे दावों के रूप में परिभाषित करते हैं, जिन्हें राज्य ने मान्य किया हो। कानूनी मान्यता से यकीनन हमारे अधिकारों को समाज में एक खास दर्जा मिलता है, लेकिन कानूनी मान्यता के आधार पर किसी अधिकार का दावा नहीं किया जा सकता।

Discussion

13.	If Dx=1 and D=1 then x=?
Answer» ANSWER: then X will be 1/1 = 1 .. .. ..

13.

If Dx=1 and D=1 then x=?

Answer»

ANSWER:

then X will be 1/1 = 1 .. .. ..

Discussion

14.	5(iii) 4.73 ஐ 4 தசம இடத் திருத்தமாக.
Answer» ANSWER: xxmjskssknlwowbw wjwiwjbwiissnd

14.

5(iii) 4.73 ஐ 4 தசம இடத் திருத்தமாக.

Answer»

ANSWER:

xxmjskssknlwowbw wjwiwjbwiissnd

Discussion

15.	the area of a trapezium is 1824 cm sq if the length of its parallel sides is 55 cm and 59 cm find the distance between them
Answer» Question : The AREA of a trapezium is 1824 cm². If the length of its parallel sides are 55 cm and 59 cm, FIND the distance between them. Answer : $\sf\pink{Given:}$ Area of Trapezium is 1824 cm² . Length of one of the parallel SIDE is 55 cm. Length of the other parallel side is 59 cm. $\sf\pink{To\:find:}$ Distance between the parallel side in short height . Explanation : $\sf\red{Area\:of\:Trapezium = \dfrac{1}{2}\times ( a + <klux>B</klux> )\times h )}$ a + b mean sum of parallel sides . $\\ \longrightarrow\sf{ 1824 = \dfrac{1}{2}\times ( 55 + 59 )\times h}$ $\\ \longrightarrow\sf{1824 = \dfrac{1}{2}\times 114\times h}$ $\\ \longrightarrow\sf{ 1824 = 57\times h}$ $\\ \longrightarrow\sf{ h = \dfrac{1824}{57}}$ $\\ \longrightarrow\sf{ h = 32}$ THEREFORE the height or the distant between the parallel sides of the Trapezium is 32 cm .

15.

the area of a trapezium is 1824 cm sq if the length of its parallel sides is 55 cm and 59 cm find the distance between them

Answer»

Question :

The AREA of a trapezium is 1824 cm². If the length of its parallel sides are 55 cm and 59 cm, FIND the distance between them.

Answer :

$\sf\pink{Given:}$

Area of Trapezium is 1824 cm² .

Length of one of the parallel SIDE is 55 cm.

Length of the other parallel side is 59 cm.

$\sf\pink{To\:find:}$

Distance between the parallel side in short height .

Explanation :

$\sf\red{Area\:of\:Trapezium = \dfrac{1}{2}\times ( a + <klux>B</klux> )\times h )}$

a + b mean sum of parallel sides .

$\\ \longrightarrow\sf{ 1824 = \dfrac{1}{2}\times ( 55 + 59 )\times h}$

$\\ \longrightarrow\sf{1824 = \dfrac{1}{2}\times 114\times h}$

$\\ \longrightarrow\sf{ 1824 = 57\times h}$

$\\ \longrightarrow\sf{ h = \dfrac{1824}{57}}$

$\\ \longrightarrow\sf{ h = 32}$

THEREFORE the height or the distant between the parallel sides of the Trapezium is 32 cm .

Discussion

16.	Total solution process see
Answer»

Discussion

17.	Find the area of triangle with vertices, P(2, 7), Q(5, 3) and R(-2, -3).
Answer» ANSWER: Area of a triangle is determinant 1/2\|x1 Y1 1 x2. Y2 1 X3 Y3 1\|

17.

Find the area of triangle with vertices, P(2, 7), Q(5, 3) and R(-2, -3).

Answer»

ANSWER:

Area of a triangle is determinant

1/2|x1 Y1 1 x2. Y2 1 X3 Y3 1|

Discussion

18.	Find the HCF of the following by continued longdivision method:(i) 60,96,150
Answer» ANSWER: use CALCULATOR Step-by-step explanation: it TAKE less TIME hjkkjhgjkfcnlafjotg

18.

Find the HCF of the following by continued longdivision method:(i) 60,96,150

Answer»

ANSWER:

use CALCULATOR

Step-by-step explanation:

it TAKE less TIME

hjkkjhgjkfcnlafjotg

Discussion

19.	(x²+ (a²+x²)½)½ differentiate wrt x
Answer» ANSWER: = -2a²x/(a²+ x²)^(2–1/2) . 1/(a² - x²)½ = -2a²x/[(a²+ x²)^3/2 .(a² - x²)½ Step-by-step EXPLANATION: -2a²x/(a²+ x²)^(2–1/2) . 1/(a² - x²)½ = -2a²x/[(a²+ x²)^3/2 .(a² - x²)½

19.

(x²+ (a²+x²)½)½ differentiate wrt x

Answer»

ANSWER:

= -2a²x/(a²+ x²)^(2–1/2) . 1/(a² - x²)½ = -2a²x/[(a²+ x²)^3/2 .(a² - x²)½

Step-by-step EXPLANATION:

-2a²x/(a²+ x²)^(2–1/2) . 1/(a² - x²)½ = -2a²x/[(a²+ x²)^3/2 .(a² - x²)½

Discussion

20.	Find the volume and surface area of the cuboid given.
Answer» HOPE it HELPED U thank u

20.

Find the volume and surface area of the cuboid given.

Answer»

HOPE it HELPED U

thank u

Discussion

21.	If F(x) = \|x\| then the range of F is equal to = a) R b) (0, ∞) c)[0, ) d) (-,0]
Answer» ANSWER: HI how are you doing TODAY

21.

If F(x) = |x| then the range of F is equal to = a) R b) (0, ∞) c)[0, ) d) (-,0]

Answer»

ANSWER:

HI how are you doing TODAY

Discussion

22.	Express each of the following as product of powers of their prime factors: 1000
Answer» Answer: 10^3 is CORRECT answer hope this will HELP

22.

Express each of the following as product of powers of their prime factors: 1000

Answer»

Answer:

10^3 is CORRECT answer

hope this will HELP

Discussion

23.	Resolve the 5x+6/2-x-x*2
Answer» Answer: B Step-by-step explanation: ONE critical point at x=1.. so TWO ranges 1) x<1.. x2−5x+6=2−−(x−1)x2−5x+6=2−−(x−1).. x2−5x+6=2+x−1x2−5x+6=2+x−1.. =>x2−6x+5=0=>x2−6x+5=0 ORx2−5x−x+5=0x2−5x−x+5=0 or (x−5)(x−1)=0(x−5)(x−1)=0 x=5 or x=1.. BUT since we are looking at the range x<1, BOTH 1 and 5 are not VALID.. 2) x>=1 x2−5x+6=2−(x−1)x2−5x+6=2−(x−1).. x2−5x+6=2−x+1x2−5x+6=2−x+1.. =>x2−4x+3=0=>x2−4x+3=0 OR x2−3x−x+3=0x2−3x−x+3=0 or(x−3)(x−1)=0(x−3)(x−1)=0 x=3 or x=1.. As we are looking at the range x>=1, BOTH 3 and 1 are VALID solution.. so product =1*3=3 B

23.

Resolve the 5x+6/2-x-x*2

Answer»

Answer:

B

Step-by-step explanation:

ONE critical point at x=1.. so TWO ranges

1) x<1..

x2−5x+6=2−−(x−1)x2−5x+6=2−−(x−1)..

x2−5x+6=2+x−1x2−5x+6=2+x−1..

=>x2−6x+5=0=>x2−6x+5=0 ORx2−5x−x+5=0x2−5x−x+5=0 or (x−5)(x−1)=0(x−5)(x−1)=0

x=5 or x=1..

BUT since we are looking at the range x<1, BOTH 1 and 5 are not VALID..

2) x>=1

x2−5x+6=2−(x−1)x2−5x+6=2−(x−1)..

x2−5x+6=2−x+1x2−5x+6=2−x+1..

=>x2−4x+3=0=>x2−4x+3=0 OR x2−3x−x+3=0x2−3x−x+3=0 or(x−3)(x−1)=0(x−3)(x−1)=0

x=3 or x=1..

As we are looking at the range x>=1, BOTH 3 and 1 are VALID solution..

so product =1*3=3

B

Discussion

24.	Class 9 maths exe 21.2 q no 1 sup no 3
Answer» WRITE down the QUESTION

Discussion

25.	the area of a Trapezium is 1824 CM square. if the length of its parallel sides are 55 cm and 59 cm find the distance between them
Answer» Answer: Length of parallel sides are 55 cm and 35 cm. Sum of parallel side's = ( 55 + 35 ) cm = 90 cm. Let DISTANCE between the two parallel sides = X cm. Area of TRAPEZIUM = 1080 cm². 1/2 × ( Sum of parallel sides ) × ( Distance between them ) = 1080 1/2 × 90 × X = 1080 90X = 2160 X = 2160/90 X = 24 cm. Hence, Distance between parallel side's = X cm = 24 cm. Step-by-step EXPLANATION:

25.

the area of a Trapezium is 1824 CM square. if the length of its parallel sides are 55 cm and 59 cm find the distance between them

Answer»

Answer:

Length of parallel sides are 55 cm and 35 cm.

Sum of parallel side's = ( 55 + 35 ) cm = 90 cm.

Let DISTANCE between the two parallel sides = X cm.

Area of TRAPEZIUM = 1080 cm².

1/2 × ( Sum of parallel sides ) × ( Distance between them ) = 1080

1/2 × 90 × X = 1080

90X = 2160

X = 2160/90

X = 24 cm.

Hence,

Distance between parallel side's = X cm = 24 cm.

Step-by-step EXPLANATION:

Discussion

26.	Find the area of the given shape.
Answer» RADIUS of HALF circle=10cm radius of FULL circle=20cm area of full circle = πr² 22/7×20×20 22/7×400 8800/7 =1257.14cm² area of semicircle =πr²/2 22/7×20×20/2 22/7×20×10 22/7×200 =4400/7 =628.5cm² area of half semicircle =628.5/2 =314.25cm² area of GIVEN figure =area of full circle - area of half semicircle 1257.14-314.25 =942.89cm²

26.

Find the area of the given shape.

Answer»

RADIUS of HALF circle=10cm

radius of FULL circle=20cm

area of full circle = πr²

22/7×20×20

22/7×400

8800/7

=1257.14cm²

area of semicircle =πr²/2

22/7×20×20/2

22/7×20×10

22/7×200

=4400/7

=628.5cm²

area of half semicircle =628.5/2

=314.25cm²

area of GIVEN figure =area of full circle - area of half semicircle

1257.14-314.25

=942.89cm²

Discussion

27.	Doll8. The shape of a garden is, as shown in the adjoiningfigure, semicircular at the ends and rectangular in themiddle. Find the area and perimeter of the garden.[HINT: Length of rectangle = 250 - 2r.]
Answer» ANSWER: SEX is the PROCESS in which MALE have FEMALE

27.

Doll8. The shape of a garden is, as shown in the adjoiningfigure, semicircular at the ends and rectangular in themiddle. Find the area and perimeter of the garden.[HINT: Length of rectangle = 250 - 2r.]

Answer»

ANSWER:

SEX is the PROCESS in which MALE have FEMALE

Discussion

28.	Cards marked with all 2digit numbers are placed in a box and are mixed throughly one card is drawn at random. Find the probability that the number on the card isa, divisble by 10b, a perfect square numberc, a prime number less than 25.
Answer» Step-by-step EXPLANATION: COME hotty rgq-dwag-dch

28.

Cards marked with all 2digit numbers are placed in a box and are mixed throughly one card is drawn at random. Find the probability that the number on the card isa, divisble by 10b, a perfect square numberc, a prime number less than 25.

Answer»

Step-by-step EXPLANATION:

COME hotty

rgq-dwag-dch

Discussion

29.	EXAMINE THE FOLLOWING TRIANGLES AND STATE THE CRITERION OF SIMILARITY SHOWN BY THEM.
Answer» ANSWER: A is the RIGHT answer....

29.

EXAMINE THE FOLLOWING TRIANGLES AND STATE THE CRITERION OF SIMILARITY SHOWN BY THEM.

Answer»

ANSWER:

A is the RIGHT answer....

Discussion

30.	•14. Multiply x2 + 7x -8 by -2y,
Answer» ANSWER: x2+7x-8by-2y = - 2x²y-14xy+16y

30.

•14. Multiply x2 + 7x -8 by -2y,

Answer»

ANSWER:

x2+7x-8by-2y

= - 2x²y-14xy+16y

Discussion

31.	25:34::76:? reasoning questionanswer plz
Answer» ANSWER: 11.18 is the answer Step-by-step EXPLANATION: because 25×34÷76

31.

25:34::76:? reasoning questionanswer plz

Answer»

ANSWER:

11.18 is the answer

Step-by-step EXPLANATION:

because 25×34÷76

Discussion

32.	The following Statement 5x+6 is called as _____ .
Answer» ANSWER: Your answer is binomial. Hope it HELPS you

32.

The following Statement 5x+6 is called as _____ .

Answer»

ANSWER:

Your answer is binomial.

Hope it HELPS you

Discussion

33.	I m not getting brainlist mark option why?????
Answer» ANSWER: I *don't* *know* Step-by-step EXPLANATION: *Please* *FOLLOW* me and MARK me as brainliest also LIKE me

33.

I m not getting brainlist mark option why?????

Answer»

ANSWER:

I don't know

Step-by-step EXPLANATION:

Please FOLLOW me and MARK me as brainliest also LIKE me

Discussion

34.	the simple interest for Rs 500 at the rate of 5% per annum is Rs 100 . The time period for the interest is ?
Answer» Step-by-step EXPLANATION: Shopix MART customer/care NUMBER 8343909193///✔️✔️✔️

34.

the simple interest for Rs 500 at the rate of 5% per annum is Rs 100 . The time period for the interest is ?

Answer»

Step-by-step EXPLANATION:

Shopix MART customer/care NUMBER 8343909193///✔️✔️✔️

Discussion

35.	Simplified from of (13)^1/5÷(13)^1/3
Answer» ANSWER: =(13)^4/15 Step-by-step EXPLANATION: 13 and13 are same so, POWER is MINUS in divide = (13)^1/3-1/5 = (13)^5-1/15 = (13)^ 4/15

35.

Simplified from of (13)^1/5÷(13)^1/3

Answer»

ANSWER:

=(13)^4/15

Step-by-step EXPLANATION:

13 and13 are same so, POWER is MINUS in divide

= (13)^1/3-1/5

= (13)^5-1/15

= (13)^ 4/15

Discussion

36.	12) A man receives 20% increase in his salary. If his newsalary is 24,600/-, then his old salary wasA) 20,000/-B) 20,500C) 21,200/-D) 21,500/-
Answer» ANSWER: dxv don't SUPPOSE wjvjdjvyeuhl B,shmtr you't

36.

12) A man receives 20% increase in his salary. If his newsalary is 24,600/-, then his old salary wasA) 20,000/-B) 20,500C) 21,200/-D) 21,500/-

Answer»

ANSWER:

dxv don't SUPPOSE wjvjdjvyeuhl B,shmtr you't

Discussion

37.	(1) By what smallest number must so bemultiplyied so that it becomeit become a perfectSanare ? Also find the square soot of thenumber so obtained.
Answer» ANSWER: YOUR ANSWER IS 56 OR YOUR ANSWER IS 65.

37.

(1) By what smallest number must so bemultiplyied so that it becomeit become a perfectSanare ? Also find the square soot of thenumber so obtained.

Answer»

ANSWER:

YOUR ANSWER IS 56

OR

YOUR ANSWER IS 65.

Discussion

38.	What is Dark Web...?
Answer» $\huge \fbox \red {A} \fbox \green {n} \fbox \purple {S} \fbox \orange {w} \fbox \pink {E} \fbox \blue {r}$ The DARK web is the World Wide Web content that exists on darknets: OVERLAY networks that use the Internet but REQUIRE specific software, CONFIGURATIONS, or authorization to ACCESS.

38.

What is Dark Web...?

Answer»

$\huge \fbox \red {A} \fbox \green {n} \fbox \purple {S} \fbox \orange {w} \fbox \pink {E} \fbox \blue {r}$

The DARK web is the World Wide Web content that exists on darknets: OVERLAY networks that use the Internet but REQUIRE specific software, CONFIGURATIONS, or authorization to ACCESS.

Discussion

39.	One angle of an isosceles triangle is 100 degree find the measure of remaining two angle
Answer» Answer: 100 + X + y = 180 x = y ( ISOSCELES PROPERTY) 100 + 2x = 180 2x = 80 x 40 So, y = 40 The other to angles are 40 degrees!

39.

One angle of an isosceles triangle is 100 degree find the measure of remaining two angle

Answer»

Answer:

100 + X + y = 180

x = y ( ISOSCELES PROPERTY)

100 + 2x = 180

2x = 80

x 40

So, y = 40

The other to angles are 40 degrees!

Discussion

40.	Find the value of k, if x-1 is a factor of 4x^3+3x^2-4x+k
Answer» $let \: p(x) = 4 {x}^{3 } + 3 {x}^{2} - 4x + k$ $if \: (x - 1 )\: is \: a \: factor \: of \: p(x) \: then \: by \: factor \: theorem \: p(1) = 0$ $p(1) = 4( {1})^{3} + 3 ({1})^{2} - 4(1) + k = 0$ $4 + 3 - 4 + k = 0$ $k = - 3$

40.

Find the value of k, if x-1 is a factor of 4x^3+3x^2-4x+k

Answer»

$let \: p(x) = 4 {x}^{3 } + 3 {x}^{2} - 4x + k$

$if \: (x - 1 )\: is \: a \: factor \: of \: p(x) \: then \: by \: factor \: theorem \: p(1) = 0$

$p(1) = 4( {1})^{3} + 3 ({1})^{2} - 4(1) + k = 0$

$4 + 3 - 4 + k = 0$

$k = - 3$

Discussion

41.	25/170 terminating or non terminating division
Answer» Answer: 25/170 terminating or NON terminating division It is a *terminating* *division* Step-by-step explanation: Thank you HOPE it HELPS you Mark me as the brainiest

41.

25/170 terminating or non terminating division

Answer»

Answer:

25/170 terminating or NON terminating division

It is a terminating division

Step-by-step explanation:

Thank you

HOPE it HELPS you

Mark me as the brainiest

Discussion

42.	Please answer fast.. find the atea of shaded region
Answer» Answer: A ( ∆ ABC ) = 1/2 × BASE × height = 1/2 × 48 × 20 [ draw a perpendicular to BC . by Pythagoras theorem we get (52)² - (48)²= (20)² . so height of triangle is 20 CM ] A ( ∆ ABC ) = 48 × 10 A (∆ ABC ) = 480 cm² now by Pythagoras theorem, in RIGHT angled ∆ ADB AD²+ DB² = AB² 12² + 16² = AB² 144+256 = AB² 400 = AB² AB = 20 cm also A( ∆ ADB ) = 1/2 × base × height = 1/2 × 16 × 12 = 16 × 6 = 96 cm² area of shaded portion = A( ∆ ABC ) - A(∆ ADB ) = 480 - 96 = 384 cm² fol low. karo yaar drop likes

42.

Please answer fast.. find the atea of shaded region

Answer»

Answer:

A ( ∆ ABC ) = 1/2 × BASE × height

= 1/2 × 48 × 20

[ draw a perpendicular to BC . by Pythagoras theorem we get (52)² - (48)²= (20)² . so height of triangle is 20 CM ]

A ( ∆ ABC ) = 48 × 10

A (∆ ABC ) = 480 cm²

now

by Pythagoras theorem,

in RIGHT angled ∆ ADB

AD²+ DB² = AB²

12² + 16² = AB²

144+256 = AB²

400 = AB²

AB = 20 cm

also

A( ∆ ADB ) = 1/2 × base × height

= 1/2 × 16 × 12

= 16 × 6

= 96 cm²

area of shaded portion = A( ∆ ABC ) - A(∆ ADB )

= 480 - 96

= 384 cm²

fol low. karo yaar

drop likes

Discussion

43.	Mr. Ram purchased 3kg tomatoes at Rs. 14.56 per kg, 3.5kg cucumber at Rs. 2.86 per kg and 16kg potatoes at Rs. 8.05per kg.a) Find the total amount spent in all.
Answer» Answer: 28.97 ANS Step-by-step explanation: 14.56+3.5+2.86+8.05 = 28.97 ans

43.

Mr. Ram purchased 3kg tomatoes at Rs. 14.56 per kg, 3.5kg cucumber at Rs. 2.86 per kg and 16kg potatoes at Rs. 8.05per kg.a) Find the total amount spent in all.

Answer»

Answer:

28.97 ANS

Step-by-step explanation:

14.56+3.5+2.86+8.05 = 28.97 ans

Discussion

44.	The value of (17)2-(17)6 is
Answer» ANSWER: -68 Step-by-step explanation: 17(2)-17(6)=34-102=-68

44.

The value of (17)2-(17)6 is

Answer»

ANSWER:

-68

Step-by-step explanation:

17(2)-17(6)=34-102=-68

Discussion

45.	In the fig., AB = CB, AB = CD and EF bisects BD at G. Prove that G is mid-point of EF.
Answer» Step-by-step explanation: Given; AB = CB AB = CD EF BISECTS BD at G. So, DG = BG; To Prove : G is the midpoint of EF or GE = GF. Construction: Join A and C with a straight line. Proof : In △ BAC; ∠CAB = ∠ACB (angles opposite to equal sides are equal) (i) In △ GDC and △GBC; GD = GB. (given) DC = BC. (given) GC = GC. (common) Therefore, △ GDC ≅ △GBC. (by SSS criteria of congruence) HENCE, ∠DGC = ∠BGC (C.P.C.T) ∠DCG = ∠BCG (C.P.C.T) (II) In △ GBC and △GBA; ∠CAB = ∠ACB (from eq. (i)) ∠AGB = ∠DGC (vertically opposute angle) AB = BC (given) Therefore, △ GBC ≅ △GBA (by AAS criteria of congruence) Hence, GC = AG (C.P.C.T) (iii) ∠BAG = ∠BCG (C.P.C.T) (iv) From equation (ii) and (iv); ∠DCG = ∠BAG. (v) ∠BAG = ∠GAF (vi) ∠DCG = ∠GCE. (VII) Now, in △CGE and △AGF; ∠GAF = ∠GCE. (from equation (v), (vi) and (vii)) AG = GC. (from equation (iii)) ∠AGF = ∠CGE. (vertically opposite angle) Therefore, △CGE ≅ △AGF (by ASA criteria of congruence) Hence, GE = GF. (by C.P.C.T) That's all.

45.

In the fig., AB = CB, AB = CD and EF bisects BD at G. Prove that G is mid-point of EF.

Answer»

Step-by-step explanation:

Given;

AB = CB

AB = CD

EF BISECTS BD at G. So, DG = BG;

To Prove : G is the midpoint of EF or GE = GF.

Construction: Join A and C with a straight line.

Proof : In △ BAC;

∠CAB = ∠ACB (angles opposite to equal sides are equal) (i)

In △ GDC and △GBC;

GD = GB. (given)

DC = BC. (given)

GC = GC. (common)

Therefore, △ GDC ≅ △GBC. (by SSS criteria of congruence)

HENCE, ∠DGC = ∠BGC (C.P.C.T)

∠DCG = ∠BCG (C.P.C.T) (II)

In △ GBC and △GBA;

∠CAB = ∠ACB (from eq. (i))

∠AGB = ∠DGC (vertically opposute angle)

AB = BC (given)

Therefore, △ GBC ≅ △GBA (by AAS criteria of congruence)

Hence, GC = AG (C.P.C.T) (iii)

∠BAG = ∠BCG (C.P.C.T) (iv)

From equation (ii) and (iv);

∠DCG = ∠BAG. (v)

∠BAG = ∠GAF (vi)

∠DCG = ∠GCE. (VII)

Now, in △CGE and △AGF;

∠GAF = ∠GCE. (from equation (v), (vi) and (vii))

AG = GC. (from equation (iii))

∠AGF = ∠CGE. (vertically opposite angle)

Therefore, △CGE ≅ △AGF (by ASA criteria of congruence)

Hence, GE = GF. (by C.P.C.T)

That's all.

Discussion

46.	sameer gave a loan of Rs 55000 to deepak for a period of one year at 16% per annum compounded half yearly. How much interest will he earn in a year
Answer» <P> Given: ✰ P = Rs 55,000 ✰ R = 16% ✰ T = 2 years ( here P = PRINCIPAL, R = rate, T = time ) To Find: • Simple INTEREST Solution: ➩ Simple Interest = Principal × Rate × Time/100 ➩ Simple Interest = P × R × T/100 Put the values in the formula, ➩ Simple Interest = 55000 × 2 × 16 100 ➩ Simple Interest = 5500 × 2 × 16 ➩ Simple Interest = 1100 × 16 ➩ Simple Interest = Rs. 17,600 ∴ SAMEER earned Interest = Rs. 17,600 Assimilate: Interest is said to be simple, if it is calculated on the ORIGINAL principal throughout the loan period. When we say simply, interest, we always mean simple interest. On the same sum and at the same rate of interest per annum, the interest of every year is same.

46.

sameer gave a loan of Rs 55000 to deepak for a period of one year at 16% per annum compounded half yearly. How much interest will he earn in a year

Answer»

<P>

Given:

✰ P = Rs 55,000

✰ R = 16%

✰ T = 2 years

( here P = PRINCIPAL, R = rate, T = time )

To Find:

• Simple INTEREST

Solution:

➩ Simple Interest = Principal × Rate × Time/100

➩ Simple Interest = P × R × T/100

Put the values in the formula,

➩ Simple Interest = 55000 × 2 × 16

100

➩ Simple Interest = 5500 × 2 × 16

➩ Simple Interest = 1100 × 16

➩ Simple Interest = Rs. 17,600

∴ SAMEER earned Interest = Rs. 17,600

Assimilate:

Interest is said to be simple, if it is calculated on the ORIGINAL principal throughout the loan period.

When we say simply, interest, we always mean simple interest.

On the same sum and at the same rate of interest per annum, the interest of every year is same.

Discussion

47.	I have 20 ,000 naira to buy shares; how many shares sold at 80k can i buy?
Answer» ANSWER: I don't understand this question. please TRY to write in detail. Step-by-step EXPLANATION: ok

47.

I have 20 ,000 naira to buy shares; how many shares sold at 80k can i buy?

Answer»

ANSWER:

I don't understand this question.

please TRY to write in detail.

Step-by-step EXPLANATION:

ok

Discussion

48.	Ir all the students of one class get together and save Rs 32 per month how many rupeswill be collected in 10 months?
Answer» Answer: YES, I THINK it is RS 320the will collect in 10 months

48.

Ir all the students of one class get together and save Rs 32 per month how many rupeswill be collected in 10 months?

Answer»

Answer:

YES, I THINK it is RS

320the will collect in 10 months

Discussion

49.	A rational number equivalent to 7/18 is
Answer» SOLUTION <P> TO DETERMINE A rational number equivalent to 7/18 is CONCEPT TO BE IMPLEMENTED RATIONAL NUMBER A Rational number is defined as a number of the form $\displaystyle \sf{ \frac{p}{q} \: }$ Where p & q are INTEGERS with $q \ne \: 0$ EXAMPLE $\displaystyle \sf{2,- 1, \frac{1}{3} , - \frac{12}{23}} \: are \: the \: examples \: of \: rational \: numbers$ EVALUATION Here the given rational number is $\displaystyle \sf{ \frac{7}{18} }$ So an rational number equivalent to given rational is $\displaystyle \sf{ \frac{7 \times 5}{18 \times 5} = \frac{35}{90} }$ FINAL ANSWER A rational number equivalent to $\displaystyle \sf{ \frac{7}{18} \: \: is \: \: \frac{35}{90} }$ ━━━━━━━━━━━━━━━━ Learn more from Brainly :- 1. Write 18.777. . . in p/q form. brainly.in/question/29915810 2. which is the following is a rational number between 4 and 8 brainly.in/question/2837211

49.

A rational number equivalent to 7/18 is

Answer»

SOLUTION

<P>

TO DETERMINE

A rational number equivalent to 7/18 is

CONCEPT TO BE IMPLEMENTED

RATIONAL NUMBER

A Rational number is defined as a number of the form $\displaystyle \sf{ \frac{p}{q} \: }$

Where p & q are INTEGERS with $q \ne \: 0$

EXAMPLE

$\displaystyle \sf{2,- 1, \frac{1}{3} , - \frac{12}{23}} \: are \: the \: examples \: of \: rational \: numbers$

EVALUATION

Here the given rational number is

$\displaystyle \sf{ \frac{7}{18} }$

So an rational number equivalent to given rational is

$\displaystyle \sf{ \frac{7 \times 5}{18 \times 5} = \frac{35}{90} }$

FINAL ANSWER

A rational number equivalent to

$\displaystyle \sf{ \frac{7}{18} \: \: is \: \: \frac{35}{90} }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. Write 18.777. . . in p/q form.

brainly.in/question/29915810

2. which is the following is a rational number between 4 and 8

brainly.in/question/2837211

Discussion

50.	In the given figure, PS and QR intersects at O and OP = OS, OQ = OR. show that ∆ POQ ≅ ∆ SOR? Also write equal pairs of angles?Pls answer both question and also give the explanation its urgent
Answer» ANSWER: In triangle POQand SoR OP=OQ(GIVEN) OQ=OR(given) angle POQ=angleSOR(vertically opp. angle) So, triangle POQ=congrurent triangle SOR(SAS) angleP=angleS(c.p.c.t)

50.

In the given figure, PS and QR intersects at O and OP = OS, OQ = OR. show that ∆ POQ ≅ ∆ SOR? Also write equal pairs of angles?Pls answer both question and also give the explanation its urgent

Answer»

ANSWER:

In triangle POQand SoR

OP=OQ(GIVEN)

OQ=OR(given)

angle POQ=angleSOR(vertically opp. angle)

So, triangle POQ=congrurent triangle SOR(SAS)

angleP=angleS(c.p.c.t)

Discussion

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