CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Take 7 to lhs and 3cotA to rhs.... take 7 common frm lhs... now multiply cotA on both sides........ then put cot sq.A= 1+cosec sq. A in lhs..... then solve it....yu will get the answer

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Prove volume of frustum

Posted by Anusha Goyal 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Siddhartha Jain 6 years, 9 months ago

Ya it is really too long aap youtube pr dekh lo

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How to convert centimetre square into litre

Posted by Shiny Thomas 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Puja Sahoo? 6 years, 9 months ago

100cm sq. = 1 litre.....

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If tanA =cot (A-18) find A

Posted by Aryan Dwivedee 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Ayushi Jain 6 years, 9 months ago

A=54

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Siddhartha Jain 6 years, 9 months ago

A=54

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if secA+tanA=p ,then prove that sinA=(p square+1)/(p square-1)

Posted by Gurpreet Kaur 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Gurpreet Kaur 6 years, 9 months ago

answer mein sinA=p sq+1/p sq-1 ana ha vo kaise

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Puja Sahoo? 6 years, 9 months ago

Multiply, ,(secA-tanA) on both sides... then solve yu will get the value of secA-tanA in terms of p... then solve the given eq. and the value yu have find to get the values of secA and tanA separately in terms of p..... then cosecA=secA/tanA put values..... yu will get cosecA= p²+1/p²-1....now, sinA=1/cosecA..... so, sinA=p sq. -1/p sq. +1

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Taking theta = 30, verify that (i) sin theta =tan theta / root 1+tan^2 theta (ii) sin 3theta = 3sin thetab- 4sin^3 theta

Posted by Friend Anjali Bisht 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Siddhartha Jain 6 years, 9 months ago

Etne simple questions to aap solve kr skte ho afterall app 10th standard m ho

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If the class marks of a continuous frequency distrubution are 22,30,38,46,54,62,...... then find the class intervals corresponding to the class markS 38 and 62

Posted by Rakesh Acharya 6 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

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Find the length of the tangent CD if ap=3cm and pc=8cm

Posted by Abhishek Abhi 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Siddhartha Jain 6 years, 9 months ago

This question is not solved without diagram

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mark a point q outside a circle c(0,2.5) such that oq is equal to 6 cm. draw two tangents to the given circle from point q.

Posted by Prasit Borah 6 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Find the largest number of 5 digits when divided by 8 & 9 leave remainder 5

Posted by Ankush Lamba 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Yashasvi Gautam 6 years, 9 months ago

99941 First subtract 5 from largest number, find the lcm of 8 & 9 divide the remaining by lcm the subtract the remainder from remaining numbet at first

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If sec0 + tan0 =p what is cosec 0 *0= theta.

Posted by Mayank Joshi 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Chinmay Phanasgaonkar 6 years, 9 months ago

sec theta+tan theta=p _________(2) sec^2 theta + tan^2 theta=1 (sec theta+tan theta)(sec theta - tan theta)= 1 sec theta- tan theta=1/sec theta + tan theta sec theta-tan theta=1/p ________(1) (1)+(2) *further you may proceed*

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If S29=99 and S99=29 find S128

Posted by Oshin Rastogi 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Oshin Rastogi 6 years, 9 months ago

Tnx but S means sum..... Solve it by taking sum of 99 terms..... ?

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Simran Kaur 6 years, 9 months ago

S29= a + 28d . S99= a + 98d .. . .a+28d=99 . a+98d=29 . ==> -70d= - 70 ===> d=1 .. .. a+28d=99 ==>a= 99-28 ==> a= 71... S128=a+127d ==) 71 + 127 ===> 198.......?

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the sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is7:15. find the numbers

Posted by Shubham Kasaudhan 6 years, 9 months ago

CBSE > Class 10 > Mathematics

4 answers

Husanpreet Singh 6 years, 9 months ago

Take four consecutive number are a-3d,a-d,a+d,a+3d

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Aliya Qureshi 6 years, 9 months ago

The numbers will be 2, 6, 10, 14

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Mayank Joshi 6 years, 9 months ago

Take 4 numbers as a-3d,a-d,a+d,a+3d.

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Honey ??? 6 years, 9 months ago

a=2 and d=4

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Tan1 tan2 tan3 tan4 ...........tan89

Posted by Shruti Choudhary 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Shruti Choudhary 6 years, 9 months ago

How mr.

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S.P Singh 6 years, 9 months ago

1Thank You

ANSWER

all 10 formula list

Posted by Amarthya Sherikar 6 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

how to solve x^2 +400x + 960000 such large quadratic equations

Posted by Amit Shinde 6 years, 9 months ago

CBSE > Class 10 > Mathematics

5 answers

Khushi ? 6 years, 9 months ago

Splitting the middle term

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✨Raagini ?? 6 years, 9 months ago

X2. + 400x - 960000 X2 + ( 1200x - 800x) + ( 1200 × 800) x ( x + 1200) - 800 ( x + 1200) (x + 1200)(x - 800)

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Ram Kushwah 6 years, 9 months ago

First find

b2-4ac=400*400-4(960000)

=160000-3840000

= negative

So the solution of this equation is not possible

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Honey ??? 6 years, 9 months ago

Take it 10^n for that

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✨Raagini ?? 6 years, 9 months ago

Use quadratic formula or completing the square

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A(-9,7)and B(4,-3) are vertices of ∆ABC . If G(1,4) is the centroid of ∆ABC , then find the third vertex.

Posted by Priyanka Singh 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Honey ??? 6 years, 9 months ago

8,8

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Prove that the line segment joining the points of contact of two parallel tangents passes through the centre

Posted by Vishakha Singhal 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Mayank Joshi 6 years, 9 months ago

Refer to rs aggarwal pg 466 theorem 10 it will help This question will be proved by the same method ??

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Hcf of 612 and 408

Posted by Akash Tiwari 6 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Yogita Ingle 6 years, 9 months ago

Prime factors of 408 = 2³ × 3 × 17 Prime factors of 612 = 2² ×3² × 17

Common prime factors: 2 , 3 , 17
HCF = 2²×3¹× 17¹ = 204

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Aliya Qureshi 6 years, 9 months ago

612=408×1+204.... 408=204×2+0........ Therefore, HCF of 612 and 408 is 204

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37x+ 43y 43x+ 37y

Posted by Karamdeep Sandhu 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Saurabh Jaiswal S.J 6 years, 9 months ago

CBSE Sample Paper for Class 11 English (Solved) – Set D

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All exercise

Posted by Hargovind Pandey 6 years, 9 months ago

CBSE > Class 10 > Mathematics

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Cards with numbers 2 to 101 are placed in a box a card is selected at random find the probability that the card have first an number second a square number

Posted by Ayushi Jain 6 years, 9 months ago

CBSE > Class 10 > Mathematics

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A contractor on construction jobs specifies a penalty for delay of completion beyond a certain date as follows rupees 200 for the first day rupees 250 for the second day ₹300 for the 3rd death penalty for each succeeding day being rupees 50 more than for the preceding day how much money the contractor has to pay as penalty if he has he did the work by 30 days

Posted by Ayushi Jain 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

S.P Singh 6 years, 9 months ago

Rs 27750

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The remainder on division of x cube + 2 X square + kx + 3 by x- 3 is 21 find the question and the value of k hence find the zeros of the cubic polynomial x cube + 2 X square + kx - 18

Posted by Ayushi Jain 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given that f(x) = x³ + 2x² + kx + 3
Let D ,Q and R are divisor,Quotient and remainder respectively
So f(x) = DQ + R
x³ + 2x² + kx + 3 = (x - 3) Q + 21

Performing long division method,

Now, we know remainder = 21
{tex}\therefore{/tex} 3 + 3(k + 15) = 21
3(k + 15) = 18
k + 15 = 6
k = -9
Hence f(x) = x³ + 2x² - 9x + 3
And Quotient = x² + 5x + (k + 15) = x² + 5x - 9 + 15 = x² + 5x + 6
Now given g(x) = x³ + 2x² + kx - 18
On substituting the value of k = -9, we get g(x) = x³ + 2x² - 9x - 18
On factorising g(x), we get,
= x² (x + 2) - 9 (x + 2)
= (x + 2)(x² - 9)
= (x + 2) (x + 3) (x - 3)
Now, g(x) = 0 if x+2=0 or x+3=0 or x-3=0
Hence, the zeros of x³ + 2x² - 9x - 18 are -2, -3 and 3.