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

we will use method of contradiction let root 2 be rational no. root2=p\q where q is not equal to zero and h.c.f.(p,q)=1 squaring on both sides, we get 2=p square÷q square 2q square=p square (1) p square is divisible by 2 therefore,p is divisible by 2 p=2k k is integer put in (1) 2q square=(2k) square 2q squrae=4k square q square = 2k square q square is divisible by 2 q is divisible by 2 now p and q both are divisible by 2 h.c.f.(p,q)=2 but their h.c.f. must be 1 so our supposition becomes wrong hence,root 2 is an irrational no. hope it helps uhh!!

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If HCF of 65 and 117 is expressible in the form of 65m-117 then find the value of m

Posted by Shineegha Thiyagarajan 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

117 = 13 {tex}\times{/tex} 3 {tex}\times{/tex} 3
65 = 13 {tex}\times{/tex} 5
HCF (117, 65) = 13
LCM(117,65) = 13 {tex}\times{/tex} 5 {tex}\times{/tex} 3 {tex}\times{/tex} 3 = 585
Here is given that:
{tex}HCF =65m-117{/tex}
13 = 65m - 117
65m = 130
m = {tex}\frac { 130 } { 6 5 } ={/tex}2

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Write all trigonometry identities?

Posted by Priyanka Chaurasiya 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Check revision notes for formulae : https://mycbseguide.com/cbse-revision-notes.html

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Find the remainder when the product of any three consecutive integers is divided by6

Posted by Tanya Khichi 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let three consecutive numbers be x, (x + 1) and (x + 2)
Let x = 6q + r 0 {tex}\leq r < 6{/tex}
{tex}\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}
{tex}\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}
if x = 6q then which is divisible by 6
{tex}\text { if } x = 6 q + 1{/tex}
{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}
{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}
{tex}= 6 ( 3 q + 1 ) \cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}
which is divisible by 6
if x = 6q + 2
{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}
{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}
{tex}= 6 ( 2 q + 1 ) \cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}
Which is divisible by 6
{tex}\text { if } x = 6 q + 3{/tex}
{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}
{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}
which is divisible by 6
{tex}\text { if } x = 6 q + 4{/tex}
{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}
{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}
which is divisible by 6
if x = 6q + 5
{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}
{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}
which is divisible by 6
{tex}\therefore {/tex} the product of any three natural numbers is divisible by 6.

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Solve (x+1)(x+2)(x+3)(x+4)=120

Posted by Ashutosh Singh 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Vivek Kohli 7 years, 6 months ago

X4 +24=120 X4=120-24 X2× x2=【4√6】2 X2=4√6 X=√4√6 2√6

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2x+3x-5

Posted by Vinit Panduru 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Vaishali Pundir 7 years, 7 months ago

By middle term spliting 2×square+(5-2)x-5 (2xsquare+5x)(2x-5) Taking x as common x(2x-5)-1 (2x-5) (x-1)(2x-5) X=1 ,X=2/5

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Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?

Posted by Ajay Pandey 7 years, 7 months ago

CBSE > Class 10 > Mathematics

2 answers

Vaishali Pundir 7 years, 7 months ago

Not 1000 its 100

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Vaishali Pundir 7 years, 7 months ago

Given in ncert Taking two variable( x-99) -(y -99)=100 _ -99 -99 cancled so x-y = 100 equation first now (X- 999)-(y-99) 999. Nd 999 cancled. So x- y =100 by euation fisrt ans is 1000

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Find least positive integer which when multiplies 9÷50 makes it a perfect square.

Posted by Neelam Ojha 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Ajay Pandey 7 years, 7 months ago

Sorry didn't get

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Any rational isintegera?

Posted by Rajnish Ojha 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

No, not every rational number is an integer.
The integers are the positive and negative whole numbers and zero: The rational numbers are numbers that can be represented by a ratio of integers.

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Out of a no. Of saras birds, one fourth are moving about on lotus plants, 1/9th coupled with the 1/4th as well as 7 times the square root of that no. Move on the hill;56 birds remain in vakula trees. What is the total no. Of saras birds?

Posted by Tanish Patial 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Siddhartha Yadav 5 years, 8 months ago

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Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes as 2,-7,-14 respectively

Posted by Vickey Gupta 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Diksha Singh 7 years, 7 months ago

Hello guys ??

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Find the no. Of divisors of 7056

Posted by Harsh Kumar 7 years, 7 months ago

CBSE > Class 10 > Mathematics

0 answers

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Ex 1.4all sums

Posted by Disha Poonja 7 years, 7 months ago

CBSE > Class 10 > Mathematics

0 answers

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Use euclid's division algorithm to find the H.C.F of 1651 and 2032. Express the H.C.F in the form of 1651m+2032 n

Posted by Harshit Srivastava 6 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Sia ? 6 years, 4 months ago

2032 = 1651 {tex} \times{/tex} 1 + 381 .
1651 = 381 {tex} \times{/tex} 4 + 127
381 = 127 {tex} \times{/tex} 3 + 0.
Since the remainder becomes 0 here, so HCF of 1651 and 2032 is 127.
{tex} \therefore{/tex} HCF (1651, 2032) = 127.
Now,
{tex} 1651 = 381 \times 4 + 127{/tex}
{tex} \Rightarrow \quad 127 = 1651 - 381 \times 4{/tex}
{tex} \Rightarrow \quad 127 = 1651 - ( 2032 - 1651 \times 1 ) \times 4{/tex} [from 2032 = 1651 {tex} \times{/tex} 1 + 381]
{tex} \Rightarrow \quad 127 = 1651 - 2032 \times 4 + 1651 \times 4{/tex}
{tex} \Rightarrow \quad 127 = 1651 \times 5 + 2032 \times ( - 4 ){/tex}
Hence, m = 5, n = -4.

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Ayush Kr Burnwal 3 years, 9 months ago

thanks a lot

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use euclid division lemma to show that the square of any positiv integer is either of the form 3m or 3m +1 for some integer m

Posted by Shivam Yadav 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let n be an arbitrary positive integer.
On dividing n by 3, let q be the quotient and r be the remainder.
Then, by Euclid's division lemma, we have
n = 3q + r, where {tex}0 \leq r < 3{/tex}.
The possibilities of remainder = 0,1 or 2
n² = (3q + r)² [∵ (a + b)² = a² + 2ab + b²]
{tex}\therefore{/tex} n² = 9q²+ r²+ 6qr ... ....(i), where {tex}0 \leq r < 3{/tex}.
Case I When r = 0.
Putting r = 0 in (i), we get
n² = 9q²
= 3(3q²)
n² = 3m, where m = 3q² is an integer.
Case II When r = 1.
Putting r = 1 in (i), we get
n² = (9q² + 1 + 6 q)
= 3(3q²+ 2q) + 1
n²= 3 m + 1, where m = (3q² + 2q) is an integer.
Case lll When r = 2.
Putting r = 2 in (i), we get
n² = (9q²+ 4 + 12q)
= 3(3q² + 4q + 1) + 1
n²= 3m + 1, where m = (3q² + 4q + 1) is an integer.
From all the above cases it is clear that the square of any positive integer is of the form 3m or (3m + 1) for some integer m.

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x + 6/y = 6 3x - 8/y =5

Posted by Riteshwar Kumar 7 years, 7 months ago

CBSE > Class 10 > Mathematics

2 answers

V Sridhar 7 years, 7 months ago

Let 1/y = a

Then, x + 6a = 6-------------------(1)
3x - 8a = 5--------------------(2)
solving the above we get a = 1/2 ; but a= 1/y . therefore y = 2
x= 3

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Anubhav . 7 years, 7 months ago

What we have to do

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Show that any positive even integer is of the form 6m,6m+2 or 6m+4,where m is some integer

Posted by Harsh Gupta 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let p be any positive integer
By division algorithm, p = 6m + r, where 0 {tex} \leqslant {/tex}r< 6

Here r=0,1,2,3,4,5
Therefore,values of p are : 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5

Now 6m+1,6m+3 and 6m+5 are odd numbers because m is a positive integer.
Hence 6m, 6m + 2, 6m + 4 are even integers because they are next positive number to the odd numbers 6m-1,6m+1 and 6m+3 respectively

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Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob age was seven times that of his son. What are their present age?

Posted by Gurdeep Maan 7 years, 7 months ago

CBSE > Class 10 > Mathematics

2 answers

V Sridhar 7 years, 7 months ago

Let the present age of Jacob = J and his son's age = S

Then, according to the problem ,
J + 5 = 3(S + 5)..............................(1)
J - 5 = 7(S - 5)................................(2)

Solving the above we get J = 40 years , S = 10 years

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Aryan Dedha 7 years, 7 months ago

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Sin60° cos30°+sin30°cos60°

Posted by Laksh Asija 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

V Sridhar 7 years, 7 months ago

We know Sin (A+B) = SinA CosB + CosA SinB

Therefore the above will be Sin (60+30) = Sin 90 = 1

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Find differentiation of sin inverse (a+bcosx/b+acosx)

Posted by Shanu Panwar 7 years, 7 months ago

CBSE > Class 10 > Mathematics

0 answers

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2+2=4why?

Posted by Mohit Mohit 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Mohit Mohit 5 years, 8 months ago

2 + X = 24 therefore X is equals to 2

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Explain why 7*6*5*4*3*2*1+5 is a composite number

Posted by Suhas Chinnu 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Vanshika Kukreja 7 years, 7 months ago

Because when we solve we see that in last we have 1009 and we know that the factor of 1009 is 1 so it is a composite no.

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What is the history of pie

Posted by Ayush Vashistha 7 years, 7 months ago

CBSE > Class 10 > Mathematics

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The sum of two number is 137 and their difference is 43 find the numbers.

Posted by Raksha Krishna Ashtankar 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Tanish Patial 7 years, 7 months ago

Let one no. Be x If the difference of the numbers is 43 then second no. Will be...x-43 ...so Acc. To question X+(x-43) = 137 2x -43 = 137 2x = 180 X=90 ->X-43=90-43=47 ;NO's are 90 and 47

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Solve by the method of completing the square X2 -6x +3=0

Posted by Nishant Gupta 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Shivam Rawat 7 years, 7 months ago

X-6x/2+3/2=0 X-3x+3/2=0 X-2(3x)/2+3/2=0 X-2(3x)/2+3/2+1-1=0 (x-1)²+3/2-1=0 (x-1)²+3/2=1 (x-1)²=1-3/2 (x-1)²=-1/2 x-1=root-,+1/2 x=1+1/2 x=3/2 When x-1=-1/2 x=-1/2+1 x=1/2 Answer is x=3/2,1/2

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If alpha and beta are the zeroes of polynomial 3x²+11x-4. Find the value of alpha/beta + beta/alpha

Posted by Shivam Rawat 7 years, 7 months ago

CBSE > Class 10 > Mathematics

0 answers

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Prove root 2 as irrational

Posted by Chinmai Urs Bhuvi 7 years, 7 months ago

CBSE > Class 10 > Mathematics

2 answers

Beauty Queen? Miss Sweetu? 7 years, 7 months ago

Here it is not possible

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Rohit Kumar 7 years, 7 months ago

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In an ap the sum of first ten terms is -80 and the sum of its next ten terms is (-280 ) find the ap

Posted by Rihan Ansari 7 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

V Sridhar 7 years, 7 months ago

The series will be 1, -1, -3, -5, -7,........

How to get it ?

The sum of n terms of an AP is given by , S = n/2 {2a+(n-1)d}

According to problem , S (10) = sum of first ten terms = -80 = 10/2 {2a + (10-1)d} = 5 (2a + 9d)------------(1)
Again sum of next 10 terms is = -280
Therefore we can write , S (20) = sum of first twenty terms = -80-280 = 20/2{20+(19-1)d} = 10(2a+19d)---------(2)

solving (1) and (2) we get a= first term = 1 and d= common difference = -2.

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Prove that is an irrational number.

Posted by Priyanka Chaurasiya 7 years, 7 months ago

CBSE > Class 10 > Mathematics