332354 + Interview Questions in Secondary School in Math

1.	What is the equation of a line passing through - 4 on the X- axis & parallel to Y-axis
Answer» ONG>ANSWER: passing through -4 PARALLEL to y- AXIS then y=0 therefore (x, y) = (-4, 0)

1.

What is the equation of a line passing through - 4 on the X- axis & parallel to Y-axis

Answer» ONG>ANSWER:

passing through -4

PARALLEL to y- AXIS then y=0

therefore (x, y) = (-4, 0)

Discussion

2.	The formula used to find the area of reactange path is
Answer» of RECTANGLE = LENGTH × breadth hope it HELPS

2.

The formula used to find the area of reactange path is

Answer»

of RECTANGLE = LENGTH × breadth

hope it HELPS

Discussion

3.	द.सा. द.शे 10%चकरवढाव्याज दराने2 वर्षात 52000 रुपयांची रास57000 रु होते तरचक्रवाढ व्याज किती?
Answer» ong>Answer: Mitochondria are membrane-bound cell organelles (mitochondrion, singular) that GENERATE most of the chemical ENERGY needed to power the cell's biochemical reactions. Chemical energy produced by the mitochondria is stored in a small MOLECULE CALLED ADENOSINE triphosphate (ATP).

3.

द.सा. द.शे 10%चकरवढाव्याज दराने2 वर्षात 52000 रुपयांची रास57000 रु होते तरचक्रवाढ व्याज किती?

Answer»

ong>Answer:

Mitochondria are membrane-bound cell organelles (mitochondrion, singular) that GENERATE most of the chemical ENERGY needed to power the cell's biochemical reactions. Chemical energy produced by the mitochondria is stored in a small MOLECULE CALLED ADENOSINE triphosphate (ATP).

Discussion

4.	2.5 is a _______ number
Answer» ONG>Answer: Decimal *PLEASE* *mark* me as *BRAINLIEST*

4.

2.5 is a _______ number

Answer» ONG>Answer:

Decimal

PLEASE mark me as BRAINLIEST

Discussion

5.	Solve 2sin^2 (x + π/3) = 1 for 0
Answer» ONG>Step-by-step EXPLANATION: $2 {sin}^{2} (<klux>X</klux> + \frac{\pi}{3}) = 1 \\ 1 - 2 {sin}^{2} (x + \frac{\pi}{3} ) = 0 \\ cos2( x + \frac{ \pi }{3} ) = 0 \\ cos(2x + \frac{2\pi}{3} ) = 0 \\ \cos(2x) \cos( \frac{2\pi}{3} ) - \sin(2x) \sin( \frac{2\pi}{3} ) = 0 \\ cos(2x) \cos( \frac{2\pi}{3} ) = \sin(2x) \sin( \frac{2\pi}{3} ) \\ \cos(2x) \cos(\pi - \frac{\pi}{3} ) = \sin(2x) \sin(\pi - \frac{\pi}{3} ) \\ \cos(2x) ( - \cos( \frac{\pi}{3} ) ) = \sin(2x) \sin( \frac{\pi}{3} ) \\ \cos(2x) ( - \frac{1}{2} ) = \sin(2x) ( \frac{ \sqrt{3} }{2}) \\ \frac{ \sin(2x) }{ \cos(2x) } = \frac{ \frac{ - 1}{2} }{ \frac{ \sqrt{3} }{2} } \\ \tan(2x) = - \frac{1}{ \sqrt{3 } } \\ \\ now \: \tan( \frac{\pi}{6} ) = \frac{1}{ \sqrt{3} } \\ \tan(2 \times \frac{\pi}{12} ) = \frac{1}{ \sqrt{3} } \\ \tan( \frac{10\pi}{12} ) = - \frac{1}{ \sqrt{3} }$ x = 10π/12

5.

Solve 2sin^2 (x + π/3) = 1 for 0

Answer» ONG>Step-by-step EXPLANATION:

$2 {sin}^{2} (<klux>X</klux> + \frac{\pi}{3}) = 1 \\ 1 - 2 {sin}^{2} (x + \frac{\pi}{3} ) = 0 \\ cos2( x + \frac{ \pi }{3} ) = 0 \\ cos(2x + \frac{2\pi}{3} ) = 0 \\ \cos(2x) \cos( \frac{2\pi}{3} ) - \sin(2x) \sin( \frac{2\pi}{3} ) = 0 \\ cos(2x) \cos( \frac{2\pi}{3} ) = \sin(2x) \sin( \frac{2\pi}{3} ) \\ \cos(2x) \cos(\pi - \frac{\pi}{3} ) = \sin(2x) \sin(\pi - \frac{\pi}{3} ) \\ \cos(2x) ( - \cos( \frac{\pi}{3} ) ) = \sin(2x) \sin( \frac{\pi}{3} ) \\ \cos(2x) ( - \frac{1}{2} ) = \sin(2x) ( \frac{ \sqrt{3} }{2}) \\ \frac{ \sin(2x) }{ \cos(2x) } = \frac{ \frac{ - 1}{2} }{ \frac{ \sqrt{3} }{2} } \\ \tan(2x) = - \frac{1}{ \sqrt{3 } } \\ \\ now \: \tan( \frac{\pi}{6} ) = \frac{1}{ \sqrt{3} } \\ \tan(2 \times \frac{\pi}{12} ) = \frac{1}{ \sqrt{3} } \\ \tan( \frac{10\pi}{12} ) = - \frac{1}{ \sqrt{3} }$

x = 10π/12

Discussion

6.	The median of 3,6,10,12,7,15,is 815
Answer» ONG>Answer: you ALREADY GAVE the answer=815 Step-by-step EXPLANATION:

6.

The median of 3,6,10,12,7,15,is 815

Answer» ONG>Answer:

you ALREADY GAVE the answer=815

Step-by-step EXPLANATION:

Discussion

7.	15% of ₹45000 How much compound interest accrued at the end of 2 years at the rate
Answer» ong>Question :- How much compound interest will be accrued at the end of 2 years at the rate of 15% PER annum for the principal of Rs 45000 compounded annually? Answer :- The compound interest accrued at the end of 2 years at the rate of 15% per annum for the principal of Rs 45000 compounded annually is Rs 14512.5 To find :- The Compound Interest. Step-by-step explanation :- In this question, the principal, rate and time has been given to US. We have to find out the compound interest. For this, first we are going to find the amount and then use it to find the compound interest. ---------------- To find the compound interest, let's find the amount first! We know that :- $\underline{\boxed{\sf Amount = Principal \Bigg(1+\dfrac{Rate}{100} \Bigg)^{Time}}}$ Here, Principal = Rs 45000. Rate = 15% per annum compounded annually. Time = 2 years. ---------------- $\underline{\underline{\mathfrak{Substituting\: the \:given \:values,}}}$ $\rightarrow\rm Amount = 45000 \Bigg(1 + \dfrac{15}{100} \Bigg)^2$ Taking 1 as the denominator and making 1 a fraction. $\rightarrow\rm Amount = \Bigg(\dfrac{1}{1} + \dfrac{15}{100} \Bigg)^2$ The LCM of 1 and 100 is 100, therefore on multiplying the fractions using their denominators, $\rightarrow\rm Amount = 45000 \Bigg(\dfrac{1 \times 100 + 15 \times 1}{100} \Bigg)^2$ On simplifying, $\rightarrow\rm Amount = 45000 \Bigg(\dfrac{100 + 15}{100} \Bigg)^2$ Adding 15 to 100, $\rightarrow\rm Amount = 45000 \Bigg(\dfrac{115}{100} \Bigg)^2$ The POWER here is 2, so REMOVING the brackets and multiplying 115/100 by itself 2 times, $\rightarrow\rm Amount = 45000 \times \dfrac{115}{100} \times \dfrac{115}{100}$ First let's multiply 115/100 with 115/100. $\rightarrow\rm Amount = 45000 \times \dfrac{115\times115}{100\times100}$ On multiplying the numbers, $\rightarrow\rm Amount = 45000 \times\dfrac{13225}{10000}$ Cutting off the zeroes, $\rightarrow\rm Amount = 45 \times \dfrac{13225}{10}$ Reducing the numbers, $\rightarrow\rm Amount = 9 \times \dfrac{13225}{2}$ Now on multiplying the remaining numbers, we get :- $\rightarrow\rm Amount = \dfrac{119025}{2}$ Dividing 119025 by 2, $\rightarrow{\overline{\boxed{\rm Amount = Rs \: 59512.5}}}$ ---------------- Now, since we know the amount, therefore let's find out the compound interest! We know that :- $\underline{\boxed{\sf CI = Amount - Principal}}$ Here, Amount = Rs 59512.5 Principal = Rs 45000. Hence, $\boxed{\bf C \: I = 59512.5 - 45000}$ Subtracting 45000 from 59512.5, $\overline{\boxed{\bf C \: I = Rs \:14512.5}}$ ---------------- Hence, the compound interest is Rs 14512.5 ----------------------------------------------------------- $\underline{\bf Abbreviation \: used :-}$ $\sf CI = Compound \: Interest.$

7.

15% of ₹45000 How much compound interest accrued at the end of 2 years at the rate

Answer»

ong>Question :-

How much compound interest will be accrued at the end of 2 years at the rate of 15% PER annum for the principal of Rs 45000 compounded annually?

Answer :-

The compound interest accrued at the end of 2 years at the rate of 15% per annum for the principal of Rs 45000 compounded annually is Rs 14512.5

To find :-

The Compound Interest.

Step-by-step explanation :-

In this question, the principal, rate and time has been given to US. We have to find out the compound interest. For this, first we are going to find the amount and then use it to find the compound interest.

----------------

To find the compound interest, let's find the amount first!

We know that :-

$\underline{\boxed{\sf Amount = Principal \Bigg(1+\dfrac{Rate}{100} \Bigg)^{Time}}}$

Here,

Principal = Rs 45000.
Rate = 15% per annum compounded annually.
Time = 2 years.

----------------

$\underline{\underline{\mathfrak{Substituting\: the \:given \:values,}}}$

$\rightarrow\rm Amount = 45000 \Bigg(1 + \dfrac{15}{100} \Bigg)^2$

Taking 1 as the denominator and making 1 a fraction.

$\rightarrow\rm Amount = \Bigg(\dfrac{1}{1} + \dfrac{15}{100} \Bigg)^2$

The LCM of 1 and 100 is 100, therefore on multiplying the fractions using their denominators,

$\rightarrow\rm Amount = 45000 \Bigg(\dfrac{1 \times 100 + 15 \times 1}{100} \Bigg)^2$

On simplifying,

$\rightarrow\rm Amount = 45000 \Bigg(\dfrac{100 + 15}{100} \Bigg)^2$

Adding 15 to 100,

$\rightarrow\rm Amount = 45000 \Bigg(\dfrac{115}{100} \Bigg)^2$

The POWER here is 2, so REMOVING the brackets and multiplying 115/100 by itself 2 times,

$\rightarrow\rm Amount = 45000 \times \dfrac{115}{100} \times \dfrac{115}{100}$

First let's multiply 115/100 with 115/100.

$\rightarrow\rm Amount = 45000 \times \dfrac{115\times115}{100\times100}$

On multiplying the numbers,

$\rightarrow\rm Amount = 45000 \times\dfrac{13225}{10000}$

Cutting off the zeroes,

$\rightarrow\rm Amount = 45 \times \dfrac{13225}{10}$

Reducing the numbers,

$\rightarrow\rm Amount = 9 \times \dfrac{13225}{2}$

Now on multiplying the remaining numbers, we get :-

$\rightarrow\rm Amount = \dfrac{119025}{2}$

Dividing 119025 by 2,

$\rightarrow{\overline{\boxed{\rm Amount = Rs \: 59512.5}}}$

----------------

Now, since we know the amount, therefore let's find out the compound interest!

We know that :-

$\underline{\boxed{\sf CI = Amount - Principal}}$

Here,

Amount = Rs 59512.5
Principal = Rs 45000.

Hence,

$\boxed{\bf C \: I = 59512.5 - 45000}$

Subtracting 45000 from 59512.5,

$\overline{\boxed{\bf C \: I = Rs \:14512.5}}$

----------------

Hence, the compound interest is Rs 14512.5

-----------------------------------------------------------

$\underline{\bf Abbreviation \: used :-}$

$\sf CI = Compound \: Interest.$

Discussion

8.	Wy) The decimal expression for 25mm in centimeters(0.25cm, 25cm, .025cm, .0025 cm)
Answer» ONG>Step-by-step EXPLANATION: The decimal expression for 25 mm = 2.5 CM 1 mm = 0.1 cm So, 25 mm = 25*0.1 cm = 2.5 cm

Discussion

9.	Decide the orderly relationship between 5/4 & 11/12 .
Answer» ONG>ANSWER: RATIO by PROPORTION is the

Discussion

10.	Anybody give me answer
Answer» ONG>ANSWER: The PICTURE is not CLEAR

Discussion

11.	Examine whether the vectors i+3j+k, 2i-j-k and 7j+5k are coplanar
Answer» ONG>ANSWER: dddvvdrbdvfbfbgnfbfbgnhmmhmh

Discussion

12.	मेवातोरडागुनवशीरिज आहे तर त्या मध्याकोलाय
Answer» ONG>ANSWER: प्रश्न का फोटो है तो पोस्ट करो, ये पढणे मे नही आ रहा

12.

मेवातोरडागुनवशीरिज आहे तर त्या मध्याकोलाय

Answer» ONG>ANSWER:

प्रश्न का फोटो है तो पोस्ट करो,

ये पढणे मे नही आ रहा

Discussion

13.	0. 646 kilo litre into litre with method answer correctly
Answer» ONG>Answer: 1 *KILOLITRE to LITRE = 1000 litre.* *therefore* , $1 \: litre \: = \frac{1}{100}$ *️‍️* $\frac{0.646}{100} litre$

13.

0. 646 kilo litre into litre with method answer correctly

Answer» ONG>Answer:

1 KILOLITRE to LITRE = 1000 litre.

therefore ,

$1 \: litre \: = \frac{1}{100}$

️‍️

$\frac{0.646}{100} litre$

Discussion

14.	Either draw a full m-ary tree with 84 leaves and height 3, where m is a positive integer, or show that no such treeexists.
Answer» TIONAL 87leves and 6height

Discussion

15.	1 point If the measure of an arc of acircle is 60° then what is themeasure of its correspondingarc?O 60O 120°O 30°O 300
Answer» ONG>Answer: 60 is the MEASURE CORRESPONDING ARC

Discussion

16.	Find the value of 1) 2 degrees minus 3 degrees 2) 2^5+2^6
Answer» ONG>ANSWER: 45 DEGREES is the CORRECT answer

16.

Find the value of 1) 2 degrees minus 3 degrees 2) 2^5+2^6

Answer» ONG>ANSWER:

45 DEGREES is the CORRECT answer

Discussion

17.	Prove thatcos 2 theta + cos² ( theta +120° ) + cos ² ( theta -120°) = ⅔
Answer» Correct Statement is $\sf \: {cos}^{2}\theta + {cos}^{2}(\theta + 120\degree) + {cos}^{2}(\theta - 120\degree) = \dfrac{3}{2}$ $\large\underline{\sf{Solution-}}$ $\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$ $\boxed{ \sf \: {cos}^{2}x = \dfrac{1 + cos2x}{2}}$ $\boxed{ \sf \: cos(x + y) + cos(x - y) = 2cosxcosy}$ $\boxed{ \sf \: cos(\pi \: + x) = - cosx}$ Let's solve the problem now!! Consider, $\sf \: {cos}^{2}\theta + {cos}^{2}(\theta + 120\degree) + {cos}^{2}(\theta - 120\degree)$ $\sf \: = \: \dfrac{1 + cos2\theta}{2} + \dfrac{1 + cos(2\theta + 240\degree)}{2} + \dfrac{1 + cos(2\theta - 240\degree)}{2}$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + cos(2\theta + 240\degree) + cos(2\theta - 240\degree) \bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + 2cos2\theta \: cos240\degree \bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + 2cos2\theta \: cos(180\degree + 60\degree) \bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta - 2cos2\theta \: cos60\degree\bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta - \cancel2cos2\theta \times \dfrac{1}{ \cancel2} \bigg)$ $\sf \: = \: \dfrac{1}{2}\bigg(3 + \cancel{ cos2\theta} - \cancel{cos2\theta} \bigg)$ $\sf \: = \: \dfrac{3}{2}$ ${\boxed{\boxed{\bf{Hence, Proved}}}}$ ─━─━─━─━─━─━─━─━─━─━─━─━─ Additional Information :- Trigonometry Formulas sin(−θ) = −sin θ cos(−θ) = cos θ tan(−θ) = −tan θ cosec(−θ) = −cosecθ sec(−θ) = sec θ cot(−θ) = −cot θ Product to Sum Formulas sin x sin y = 1/2 [cos(x–y) − cos(x+y)] cos x cos y = 1/2[cos(x–y) + cos(x+y)] sin x cos y = 1/2[sin(x+y) + sin(x−y)] cos x sin y = 1/2[sin(x+y) – sin(x−y)] Sum to Product Formulas sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2] sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2] cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2] cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2] Sum or Difference of angles cos (A + B) = cos A cos B – sin A sin B cos (A – B) = cos A cos B + sin A sin B sin (A+B) = sin A cos B + cos A sin B sin (A -B) = sin A cos B – cos A sin B tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)] tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)] cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)] cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)] cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A Multiple and Submultiple angles sin2A = 2SINA cosA = [2tan A /(1+tan²A)] cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)] tan 2A = (2 tan A)/(1-tan²A)

17.

Prove thatcos 2 theta + cos² ( theta +120° ) + cos ² ( theta -120°) = ⅔

Answer»

Correct Statement is

$\sf \: {cos}^{2}\theta + {cos}^{2}(\theta + 120\degree) + {cos}^{2}(\theta - 120\degree) = \dfrac{3}{2}$

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

$\boxed{ \sf \: {cos}^{2}x = \dfrac{1 + cos2x}{2}}$

$\boxed{ \sf \: cos(x + y) + cos(x - y) = 2cosxcosy}$

$\boxed{ \sf \: cos(\pi \: + x) = - cosx}$

Let's solve the problem now!!

Consider,

$\sf \: {cos}^{2}\theta + {cos}^{2}(\theta + 120\degree) + {cos}^{2}(\theta - 120\degree)$

$\sf \: = \: \dfrac{1 + cos2\theta}{2} + \dfrac{1 + cos(2\theta + 240\degree)}{2} + \dfrac{1 + cos(2\theta - 240\degree)}{2}$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + cos(2\theta + 240\degree) + cos(2\theta - 240\degree) \bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + 2cos2\theta \: cos240\degree \bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta + 2cos2\theta \: cos(180\degree + 60\degree) \bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta - 2cos2\theta \: cos60\degree\bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + cos2\theta - \cancel2cos2\theta \times \dfrac{1}{ \cancel2} \bigg)$

$\sf \: = \: \dfrac{1}{2}\bigg(3 + \cancel{ cos2\theta} - \cancel{cos2\theta} \bigg)$

$\sf \: = \: \dfrac{3}{2}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

Additional Information :-

Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2SINA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)

Discussion

18.	2.5 is a ______ number
Answer» ONG>ANSWER: DECIMAL Step-by-step explanation: 2.5 is a decimal number. Hope it HELPS

18.

2.5 is a ______ number

Answer» ONG>ANSWER:

DECIMAL

Step-by-step explanation:

2.5 is a decimal number.

Hope it HELPS

Discussion

19.	What is the compound Interest on le 200for Iyear at 10% reale compenyearly
Answer» ONG>Answer: 20 Step-by-step EXPLANATION: ??????????????????

19.

What is the compound Interest on le 200for Iyear at 10% reale compenyearly

Answer» ONG>Answer:

20

Step-by-step EXPLANATION:

??????????????????

Discussion

20.	13cm 8. a) The adjoining solid object is formed with the combination of apyramid and a square-based prism. If the volume of the solidobject is 900 cu.cm, find the height of the prism.10 cm
Answer» ONG>ANSWER: hkkhtrvboourdvjiijbfewhj

Discussion

21.	What is the equation of a line Passing through - 4the x-axis& Parallel to Y - axis ?
Answer» Line passing through(-4,0) and parallel to y-axis. So, equation of line X = -4 PLS pls MARK BRAINLIEST!!

21.

What is the equation of a line Passing through - 4the x-axis& Parallel to Y - axis ?

Answer»

Line passing through(-4,0) and parallel to y-axis.

So, equation of line X = -4

PLS pls MARK BRAINLIEST!!

Discussion

22.	1. Convert the following into hours.al 3 days
Answer» ONG>ANSWER: 72 HOURS is the CORRECT answer

Discussion

23.	The common factor of the algebraic expressions ax2y,bxy2 and cxyz is
Answer» s the COMMON factor for the algebraic expression. if we take out xy we GET a2, B2,CZ in which there's NOTHING common.

23.

The common factor of the algebraic expressions ax2y,bxy2 and cxyz is

Answer»

s the COMMON factor for the algebraic expression.

if we take out xy

we GET a2, B2,CZ in which there's NOTHING common.

Discussion

24.	₹45000 पर 15% वर्सिक। दर से 2 वर्ष के अंत में कितना चक्रवृधि ब्याज उपचित होगा
Answer» ONG>ANSWER: SORRY I can't understand HINDI so can you write in English

Discussion

25.	draw a draw a a diagram draw a diagram a cubical block of side 10cm surrounded by a hemisphereड्रॉ ए डायग्राम ड्रॉ ए क्यूबिकल ब्लॉक आफ साइड 10 सेंटीमीटर सर माउंटेड बाय हैव इस पर व्हाट इज द लार्जेस्ट डायमीटर ऑफ द ईयर रुपीस 5 स्क्वायर मीटर
Answer» ANSWER STEP to step EXPLANATION:- HOPE its HELPFUL to you .

25.

draw a draw a a diagram draw a diagram a cubical block of side 10cm surrounded by a hemisphereड्रॉ ए डायग्राम ड्रॉ ए क्यूबिकल ब्लॉक आफ साइड 10 सेंटीमीटर सर माउंटेड बाय हैव इस पर व्हाट इज द लार्जेस्ट डायमीटर ऑफ द ईयर रुपीस 5 स्क्वायर मीटर

Answer»

ANSWER

STEP to step EXPLANATION:-

HOPE its HELPFUL to you .

Discussion

26.	Find the value of y in the following equation 2x-1/2 +1 - 3xy/4
Answer» ong>Answer: Hi ! LET 1/3x + y = a , and 1/3x - y = b , a + b = 3/4 -----------------> (1) 4A + 4b = 3 -------------> (1) also , a/2 - b/2 = -1/8 (a - b)/2 = -1/8 a - b = -1/4 4a - 4b = -1 ---------------------------> (2) Adding EQUATIONS (1) and (2) :- 4a + 4b = 3 4a - 4b = -1 -------------------- 8a = 2 a = 2/8 a = 1/4 4a - 4b = -1 *41/4 - 4b = -1 1 - 4b = -1 4b = 1 + 1 4b = 2 b = 2/4 b = 1/2 1/3x + y = a 1/3x + y = 1/4 3x + y = 4 -----------------> (3) 1/3x - y = b 1/3x - y = 1/2 3x - y = 2 -----------------> (2) Adding equations (3) and (4) :- 3x + y = 4 3x - y = 2 -------------------------- 6X = 6 x = 1 3x + y = 4 31 + y = 4* y = 4 - 3 = 1 Hence, x = 1 , y = 1

26.

Find the value of y in the following equation 2x-1/2 +1 - 3xy/4

Answer»

ong>Answer:

Hi !

LET 1/3x + y = a , and 1/3x - y = b ,

a + b = 3/4 -----------------> (1)

4A + 4b = 3 -------------> (1)

also ,

a/2 - b/2 = -1/8

(a - b)/2 = -1/8

a - b = -1/4

4a - 4b = -1 ---------------------------> (2)

Adding EQUATIONS (1) and (2) :-

4a + 4b = 3

4a - 4b = -1

--------------------

8a = 2

a = 2/8

a = 1/4

4a - 4b = -1

4*1/4 - 4b = -1

1 - 4b = -1

4b = 1 + 1

4b = 2

b = 2/4

b = 1/2

1/3x + y = a

1/3x + y = 1/4

3x + y = 4 -----------------> (3)

1/3x - y = b

1/3x - y = 1/2

3x - y = 2 -----------------> (2)

Adding equations (3) and (4) :-

3x + y = 4

3x - y = 2

--------------------------

6X = 6

x = 1

3x + y = 4

3*1 + y = 4

y = 4 - 3 = 1

Hence,

x = 1 , y = 1

Discussion

27.	P(x) = x2-3x+2q(x) = 3x+1(i) find p(x)+q(x)(ii) what number is p(x)+ 2(x)
Answer» ONG>ANSWER: ESKA answer upper PAPER me hai.

27.

P(x) = x2-3x+2q(x) = 3x+1(i) find p(x)+q(x)(ii) what number is p(x)+ 2(x)

Answer» ONG>ANSWER:

ESKA answer upper PAPER me hai.

Discussion

28.	Substract 14.256 from 27.830
Answer» ONG>ANSWER: Hii.Mark me as the BRAINLIST

Discussion

29.	How to know how many thousands in 1780535
Answer» CCcbcd MEIN BHI EK HI

Discussion

30.	Find the volume and total surface area of a cylinder whose ra- dius of the base is 21 cm ad height is 50 cm.
Answer» ONG>Step-by-step explanation: Here, RADIUS = 21 cm Height = 50 cm So, VOLUME of CYLINDER = πr^2h = 22/7212150 = 2232150 = 69300 cm^3 Now, Total surface area of cylinder = 2πr(r+h) = 222/721(21+50) = 44371 = 9372 cm^2

Discussion

31.	What is the compound Interest on Rs 2012. at 10% rate Compound halteyearlyfor year
Answer» ong>Step-by-step explanation: P=Rs. 10000 T=12mon =1 year R=10% p.a. Compounded HALF yearly n=2 ∴A=P[1+ n/r]NNT =10000[1+2/10]2×1 =10000[2/12]2 =10000×6×6 A=360000 C.I=A−P =Rs. (360000−10000) =Rs. 260000. hope it helps

31.

What is the compound Interest on Rs 2012. at 10% rate Compound halteyearlyfor year

Answer»

ong>Step-by-step explanation:

P=Rs. 10000

T=12mon =1 year

R=10% p.a.

Compounded HALF yearly

n=2

∴A=P[1+ n/r]NNT

=10000[1+2/10]2×1

=10000[2/12]2

=10000×6×6

A=360000

C.I=A−P

=Rs. (360000−10000)

=Rs. 260000.

hope it helps

Discussion

32.	1.6 x 2
Answer» ONG>Step-by-step explanation: 13²=169 160=(13-Δ)²=169–26Δ+Δ² 9≈26Δ Δ≈9/26≈0.346 √160≈13–0.346≈12.65 √1.6≈1.265

32.

1.6 x 2

Answer» ONG>Step-by-step explanation:

13²=169

160=(13-Δ)²=169–26Δ+Δ²

9≈26Δ

Δ≈9/26≈0.346

√160≈13–0.346≈12.65

√1.6≈1.265

Discussion

33.	33. From the top of a building 20 meter high, the angle of elevation of the top of a vertical pole is 30°, and the angle of depression of the foot of the same poleis 60°. Find the height of the pole.
Answer» ER: the ANSWER is 40M.

Discussion

34.	Solve it without taking it common 1290=P2323
Answer» ONG>Step-by-step EXPLANATION: this is the SOLUTION of the QUESTION.... dgdvxjdb.. udhdn

34.

Solve it without taking it common 1290=P2323

Answer» ONG>Step-by-step EXPLANATION:

this is the SOLUTION of the QUESTION....

dgdvxjdb..

udhdn

Discussion

35.	1/1+cot²theta =___---_
Answer» now it is PROBE wala question h ISKE AAGE wala bhejo

35.

1/1+cot²theta =___---_

Answer»

now it is PROBE wala question h ISKE AAGE wala bhejo

Discussion

36.	4 the correct alternatives of the following1) The areas of two similar triangle are 36cm and 8cm. The ratio of theircorresponding height isa) 9:16 b) 6:9 c) 36:81
Answer» ong>Answer: 6 : 9 (Option b) Step-by-step EXPLANATION: Let the areas of the two triangles be A₁ and A₂. Let the heights of the two triangles be h₁ and h₂. Let ONE side of the first triangle be s₁ and its corresponding side in the second triangle be s₂. Given, ratios of areas of triangles = A₁ : A₂ = 36 : 81 We know that the areas of SIMILAR triangles are in the same ratio as the squares of their corresponding sides. So, A₁ : A₂ = s₁² : s₂² = 36 : 81 --------> ( I ) We know that the corresponding sides of similar triangles are in the same ratio as their corresponding heights. So, s₁ : s₂ = h₁ : h₂ --------------------------> ( II ) From ( I ), ( II ), s₁² : s₂² = h₁² : h₂² = 36 : 81 h₁² : h₂² = 36 : 81 h₁ : h₂ = √36 : √81 = 6 : 9 ∴ Ratio of corresponding heights of triangles = 6 : 9 (Option b) Hope it helps!!! PLEASE MARK Brainliest!!!

36.

4 the correct alternatives of the following1) The areas of two similar triangle are 36cm and 8cm. The ratio of theircorresponding height isa) 9:16 b) 6:9 c) 36:81

Answer»

ong>Answer:

6 : 9 (Option b)

Step-by-step EXPLANATION:

Let the areas of the two triangles be A₁ and A₂.

Let the heights of the two triangles be h₁ and h₂.

Let ONE side of the first triangle be s₁ and its corresponding side in the second triangle be s₂.

Given, ratios of areas of triangles = A₁ : A₂ = 36 : 81

We know that the areas of SIMILAR triangles are in the same ratio as the squares of their corresponding sides.

So,

A₁ : A₂ = s₁² : s₂² = 36 : 81 --------> ( I )

We know that the corresponding sides of similar triangles are in the same ratio as their corresponding heights.

So,

s₁ : s₂ = h₁ : h₂ --------------------------> ( II )

From ( I ), ( II ),

s₁² : s₂² = h₁² : h₂² = 36 : 81

h₁² : h₂² = 36 : 81

h₁ : h₂ = √36 : √81 = 6 : 9

∴ Ratio of corresponding heights of triangles = 6 : 9 (Option b)

Hope it helps!!! PLEASE MARK Brainliest!!!

Discussion

37.	What sum of money at compound interest will amount to Rs. 6930 in 3 years, if the rate of interest is the first year, 10% for the second year and 20% for the third year?
Answer» ONG>Answer: this FINAL amount will be 7257 after 3 years at 20

Discussion

38.	The place value of 3 in 6.053 ___a) Tenths b) Hundredthsc) Thousandthsd) Hundreds
Answer» ER: OPTION C Step-by-step EXPLANATION: Thousandths Hope it HELPS

38.

The place value of 3 in 6.053 ___a) Tenths b) Hundredthsc) Thousandthsd) Hundreds

Answer» ER:

OPTION C

Step-by-step EXPLANATION:

Thousandths

Hope it HELPS

Discussion

39.	Cot theta/cosectheta
Answer» ONG>Step-by-step explanation: Tan (THETA) = Opposite/Adjacent. Cot (theta) = Adjacent/Opposite. Cosec (theta) = Hypotenuse/Opposite. Sec (theta) = Hypotenuse/Adjacent. BY TEJAS KULKARNI PLEASE FOLLOW AND MARK AS BRAINLIST AND FOLLOW FOR MORE ANSWERS!!!

39.

Cot theta/cosectheta

Answer» ONG>Step-by-step explanation:

Tan (THETA) = Opposite/Adjacent. Cot (theta) = Adjacent/Opposite. Cosec (theta) = Hypotenuse/Opposite. Sec (theta) = Hypotenuse/Adjacent.

BY TEJAS KULKARNI PLEASE FOLLOW AND MARK AS BRAINLIST AND FOLLOW FOR MORE ANSWERS!!!

Discussion

40.	The sum of five terms of an arithmetic sequence is 60. a) find the middle term.b) write three arithmetic sequence whose sum of first five terms is 60.
Answer» ONG>Answer: 12 + 12 + 12 + 12 + 12 +12 = 6O Middle TERM 12

40.

The sum of five terms of an arithmetic sequence is 60. a) find the middle term.b) write three arithmetic sequence whose sum of first five terms is 60.

Answer» ONG>Answer:

12 + 12 + 12 + 12 + 12 +12 = 6O

Middle TERM 12

Discussion

41.	The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. Then calculate ratio of the surface area of the balloon inthe two case.
Answer» ONG>ANSWER: the answer of this QUESTION is 72

Discussion

42.	3. fond three Consecutive Numbes whose sum.ismare than 55 and less than 60
Answer» ong>Answer: The three consecutive numbers are 18, 19, 20 Step-by-step explanation: 18 + 19 + 20 = 57 57 > 55 and 57 < 60

42.

3. fond three Consecutive Numbes whose sum.ismare than 55 and less than 60

Answer»

ong>Answer:

The three consecutive numbers are 18, 19, 20

Step-by-step explanation:

18 + 19 + 20 = 57

57 > 55 and 57 < 60

Discussion

43.	A certain sum amounts to INR 9799 in 3 years and to INR 12464 in 8 years at a certain rate percent per annum, at simple interest. What will be the amount (in rupees) of the same sum in 5 years at double therate, at simple interest?O 13530O 12840O 12960O 13500
Answer» ong>Step-by-step explanation: PLEASE give me THANKS and MARK me brilient ☺️

43.

A certain sum amounts to INR 9799 in 3 years and to INR 12464 in 8 years at a certain rate percent per annum, at simple interest. What will be the amount (in rupees) of the same sum in 5 years at double therate, at simple interest?O 13530O 12840O 12960O 13500

Answer»

ong>Step-by-step explanation:

PLEASE give me THANKS and MARK me brilient ☺️

Discussion

44.	(2) Form a two digit number whose unit's place is x and ten's place is'y.
Answer» + x hope it HELPS you BRO

44.

(2) Form a two digit number whose unit's place is x and ten's place is'y.

Answer»

+ x

hope it HELPS you BRO

Discussion

45.	at present a mother is twice as old as his daughter. twenty years back their ages were in the ratio of 10 : 3 .The present age of the mother is
Answer» ong>Answer: the present age of the mother is 70 i hope this helps!! Step-by-step explanation: let us take the present age of the mother as x and the present age of her daughter as y "at present a mother is twice as old as her daughter" so, we get the first equation as $x = 2y$ from this. "twenty years back their ages" Now, 20 years ago, their ages will be 20 years less., so, the past age of the mother is 20-x and the past age of the daughter is 20-y "twenty years back their ages were in the ratio of 10 : 3" since ratio is a fraction, we can SAY that $\frac{20-x}{20-y} = \frac{10}{3}$ substituting the $x$ with the first equation so we get: $\frac{20-(2y)}{20-y} = \frac{10}{3}$ now we cross multiply, which gives us, => $3(20-2y) = 10(20-y)$ => $60 - 6y = 200 - 10y$ => $- 6y + 10y = 200 - 60$ => $4y = 140$ => $y = \frac{140}{4}$ => $y = 30$ to find the mothers present age, we USE the first equation; $x = 2y$ $x = 2$ × $35$ $x = 70$

45.

at present a mother is twice as old as his daughter. twenty years back their ages were in the ratio of 10 : 3 .The present age of the mother is

Answer»

ong>Answer:

the present age of the mother is 70

i hope this helps!!

Step-by-step explanation:

let us take the present age of the mother as x and the present age of her daughter as y

"at present a mother is twice as old as her daughter" so,
we get the first equation as $x = 2y$ from this.
"twenty years back their ages" Now, 20 years ago, their ages will be 20 years less., so, the past age of the mother is 20-x and the past age of the daughter is 20-y
"twenty years back their ages were in the ratio of 10 : 3" since ratio is a fraction, we can SAY that $\frac{20-x}{20-y} = \frac{10}{3}$

substituting the $x$ with the first equation so we get:

$\frac{20-(2y)}{20-y} = \frac{10}{3}$

now we cross multiply, which gives us,

=> $3(20-2y) = 10(20-y)$

=> $60 - 6y = 200 - 10y$

=> $- 6y + 10y = 200 - 60$

=> $4y = 140$

=> $y = \frac{140}{4}$

=> $y = 30$

to find the mothers present age, we USE the first equation;

$x = 2y$

$x = 2$ × $35$

$x = 70$

Discussion

46.	If nth term of GP is (-3)^n÷2^n-1 then common ratio is207
Answer» ONG>Step-by-step EXPLANATION: an+1=(-3)^n+1/2^n an=(-3)^n/2^n-1 common ratio=an+1/an =(-3)^n+1/(-3)^n. (2^n-1/2^n) =(-3)^n+1-n (2)^n-1-n =-3/2

46.

If nth term of GP is (-3)^n÷2^n-1 then common ratio is207

Answer» ONG>Step-by-step EXPLANATION:

an+1=(-3)^n+1/2^n

an=(-3)^n/2^n-1

common ratio=an+1/an

=(-3)^n+1/(-3)^n. *(2^n-1/2^n)

=(-3)^n+1-n *(2)^n-1-n

=-3/2

Discussion

47.	16 : 24 :: 20 : 30 are in proportion TRUE OR False
Answer» ONG>Answer: 16 : 24 :: 20 : 30 are in PROPORTION *TRUE* OR False Step-by-step explanation: It is true dear. pls amrk me as BRAINLIEST and give me thanks.

47.

16 : 24 :: 20 : 30 are in proportion TRUE OR False

Answer» ONG>Answer:

16 : 24 :: 20 : 30 are in PROPORTION TRUE OR False

Step-by-step explanation:

It is true dear.

pls amrk me as BRAINLIEST and give me thanks.

Discussion

48.	Kavin scored 15 out of 25 in test the percentage of his mark is __
Answer» ong>Answer: *kavin scored 15 out of 25 in TEST the PERCENTAGE of his mark is* 60° Step-by-step explanation: please mark it brainliest and folow me

48.

Kavin scored 15 out of 25 in test the percentage of his mark is __

Answer»

ong>Answer:

kavin scored 15 out of 25 in TEST the PERCENTAGE of his mark is 60°

Step-by-step explanation:

please mark it brainliest and folow me

Discussion

49.	If the roots of the equations (a-b)x2+(b-c)x+(c-a)=0
Answer» ong>Answer: 2=b+c Step-by-step explanation: The question MUST be, if ROOTS of equation (a-b)x2+(b-c)x+(c-a)=0 are equal , prove that b+c = 2a. Using Discriminant, D = B2-4AC as compared with the general quadratic equation Ax2+Bx+C=0 so, A = a-b B = b-c C = c-a For roots to be equal, D=0 (b-c)2 - 4(a-b)(c-a) =0 b2+c2-2bc -4(ac-a2-bc+ab) =0 b2+c2-2bc -4ac+4a2+4bc-4ab=0 4a2+b2+c2+2bc-4ab-4ac=0 (2a-b-c)2=0 i.e. 2a-b-c =0 2a= b+c

49.

If the roots of the equations (a-b)x2+(b-c)x+(c-a)=0

Answer»

ong>Answer:

2=b+c

Step-by-step explanation:

The question MUST be, if ROOTS of equation (a-b)x2+(b-c)x+(c-a)=0 are equal , prove that b+c = 2a.

Using Discriminant,

D = B2-4AC as compared with the general quadratic equation Ax2+Bx+C=0

so, A = a-b

B = b-c

C = c-a

For roots to be equal, D=0

(b-c)2 - 4(a-b)(c-a) =0

b2+c2-2bc -4(ac-a2-bc+ab) =0

b2+c2-2bc -4ac+4a2+4bc-4ab=0

4a2+b2+c2+2bc-4ab-4ac=0

(2a-b-c)2=0

i.e. 2a-b-c =0

2a= b+c

Discussion

50.	Factorise in grouping method a^4-2a^3-4a+8.
Answer» ong>Step-by-step EXPLANATION: i HOPE its helpful and UNDERSTAND for U

50.

Factorise in grouping method a^4-2a^3-4a+8.

Answer»

ong>Step-by-step EXPLANATION:

i HOPE its helpful and UNDERSTAND for U

Discussion

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