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Amit Kumar 9 hours ago

Trigonometry

Aman Choudhary 10 hours ago

Yes

Yogita Ingle 15 hours ago

2x + 3y = 7
2ax + (a + b) y = 28
If two equations has infinite solutions, then,
a1 / a2 = b1 / b2 = c1 /c2
2 / 2a = 3 / a+b = -7 / -28
1/a = 3/a+b = 1/4
Now,
1/a = 1/4
a = 4
And
3/a+b = 1/4
12 = 4 + b
12 - 4 = b
b = 8
Thus,
a = 4
b = 8

So, b = 2a

Yogita Ingle 15 hours ago

( 17 x 11 x 2 ) + ( 17 x 11 x 5 )
By taking ( 17 x 11 ) common ,
= ( 17 x 11 ) ( 2 + 5 )
= ( 17 x 11 ) ( 7 )
= 17 x 11 x 7.
If any number can be represented in the form of product of prime numbers then that number is composite number.
Hence , ( 17 x 11 x 2 ) + ( 17 x 11 x 5 ) is a composite number.

Shambhu Shind 9 hours ago

Prime factors of 378=2*3*3*3*7 Prime factors of 180=2*2*3*3*5 Prime factors of 420=2*2*3*5*7 So, hcf of 378,180,420 is 2*3=6 LCM of 348,180,420 is 2*2*3*3*3*5*7=3780

Gaurav Seth 18 hours ago

A n s w e r

Prime factors of 378 = 2 × 3 × 3 × 3 × 7
Prime factors of 180 = 2 × 2 × 3 × 3 × 5
Prime factors of 420 = 2 × 2 × 3 × 5 × 7
So, HCF of 378, 180 and 420 is 2 × 3 = 6
And LCM of 378, 180 and 420 is 2 × 2 × 3 × 3 × 3 × 5 × 7 = 3780

Yogita Ingle 21 hours ago

Given, $P(E) = 0.05$

We know,

$P(not\ E) = 1 - P(E)$

$\therefore P(not\ E) = 1 - 0.05 = 0.95$

Hence, the probability of 'not E' is 0.95

Ni Nanthi 22 hours ago

7+7=14

Gaurav Seth 23 hours ago

1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

Solution:

(i)Let there are x number of girls and y number of boys. As per the given question, the algebraic expression can be represented as follows.

x +y = 10

x– y = 4

Now, for x+y = 10 or x = 10−y, the solutions are;

For x – y = 4 or x = 4 + y, the solutions are;

The graphical representation is as follows;

Click on the given link to continue

### <a href="https://mycbseguide.com/blog/ncert-solutions-for-class-10-maths-exercise-3-2/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-for-class-10-maths-exercise-3-2/&ved=2ahUKEwj2r_u647_sAhWOwjgGHT1jADUQFjAEegQIBBAC" rel="noopener" target="_blank">NCERT Solutions for Class 10 Maths Exercise 3.2</a>

Sourav Keshri 23 hours ago

Hai bhai
I am the student of class 10 I am on Telegram
Hiii bro

Sourav Keshri 23 hours ago

X =(2+15y)/7 Y =19/17

Shubham Kumar 1 day, 12 hours ago

Here your answer goes Step :-1 Take one equation from the two equations and represent it in the form of x and y by replacing 7x - 15y = 2----- > 1 x + 2y = 3------- > 2 Take equation 2 x = 3 - 2y ------>> 3 Step :- 2 Now , Substitute the value of x in equation 1 7 ( 3 - 2y ) - 15 = 2 21 - 14y - 15y = 2 21 - 29y = 2 -29y = 21 - 2 y = 29/19 ​ Step :- 3 Put the obtained value of y in equation 3 x = 3 - 2yx=3−2y x = 29/49 Hope you like this answer ​

Sourav Keshri 23 hours ago

Sin square 60°+2tan 45°- cos square 30° (√ 3/4)square + 2x1- (√ 3/4)square 3/4 + 2/1 -3/4 (+3/4 or -3/4 cut ho jayega) +2/1 = (2 Ans)

Sourav Keshri 22 hours ago

316

Yogita Ingle 1 day, 13 hours ago

We know that the largest 5 digit number is 99999.

So here we can see that 99999 is not a perfect square as it leaves 143 as the remainder while finding it s square root.

So for finding 5 digit perfect square we will have to subtract 143 from 99999

=> 99999-143 = 99856

Hence 99856 is a perfect square whose square root is √99856 = 316.

Arpit Punia 1 day, 15 hours ago

(3) Disease

Sourav Keshri 22 hours ago

(Ans) :- X square + 3x + 4

Sumit Choudhury 1 day, 16 hours ago

x^2-x-12=0

Yogita Ingle 1 day, 12 hours ago

we know that if there is no solution then lines must be parallel and if lines are parallel than

≠ c1/c2

so,a1/a2=3/6= 1/2

b1/b2=1 / k

now

so k=2

the square of the hypotenuse side is equal to the sum of squares of the other two sides is know as phythagoras theorem Formula : Hypotenuse2 = Perpendicular2 + Base2  c2 = a2 + b2

Jobanjot Kaur 1 day, 21 hours ago

The square of two the hypotenuse is the sum of the other two sides is known as pythagoras theorem FORMULA: H2 =B2+P2 H Square =B Square +P Square

Shuvendu Kumar 1 day, 21 hours ago

The square of two the hypotenuse is the sum of other two sides is known as pythagoras therom

Harshali Vilhekar 2 days, 7 hours ago

The square of the hypotenuse is equal to the sum of squares of other two sides is known as Pythagoras theorem

Gaurav Seth 2 days, 16 hours ago

In figure, a circle touches all the four sides of a quadrilateral ABCD/ whose sides AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

<hr />

Since, length of two tangents drawn from an external point of circle are equal.
So,    AP = AS
BP = BQ DR = DS
and    RC = CQ
(AP + BP) + (DR + RC) = AS + BQ + DS + CQ
⇒ (AP + BP) + (DR + RC)
= (AS + DS) + (BQ + CQ)
⇒    AB + DC = AD + BC
⇒    6 + 4 = AD + 7
⇒    10 = AD + 7
AD = 10 – 7 = 3 cm

Ekta Sirohi 1 day, 15 hours ago

Good afternoon

😉😉 😝😝 1 day, 19 hours ago

Good afternoon.

Nisha😋🤣 Bharti 2 days, 18 hours ago

Good afternoon

Chirag Sharma 2 days, 17 hours ago

SecA= 4√7/7

Bhai Se Sikho Sikho 2 days, 17 hours ago

sin A = 3/4 P = 3 H= 4 B=? H² = P² + B² B²= 16 - 9 B = √7 Sec A = H/B = 4/√7 4/√7 × √7/√7 = 4√7/7 So, Sec A = 4√7/7

Nisha😋🤣 Bharti 2 days, 18 hours ago

Sin A= 3/2=p/h b²=h²-p²= b²=4-9=-5 b=√-5 SecA=h/b=2/-√5

Maniya Jain 2 days, 19 hours ago

Sec A = h / b = 4/√7

Taniya Chouhan 2 days, 21 hours ago

secA = 4/√7

Ekta Sirohi 1 day, 14 hours ago

LCM(6,20)= 60
60

Yogita Ingle 2 days, 23 hours ago

6 = 21 × 31

20 = 22 × 51

LCM(6,20) = 22 × 31 × 51
LCM(6,20) = 60

Yogita Ingle 2 days, 23 hours ago

The common point of a tangent to a circle and the circle is called___

Point of contact

Sourav Keshri 22 hours ago

Value of α2+ β2.

Yogita Ingle 2 days, 23 hours ago

Value of α2+ β2

Maniya Jain 2 days, 19 hours ago

2x-y=2 2x-2=y ....(1) x+3y=15 ....(2) By substituting the value from 1 and 2 eq. x+3(2x-2)=15 x+6x-6=15 7x-6=15 7x = 21 x= 21÷7 x=3 From 1 eq we can find out the value of y ..... 2x-2=y 2(3)-2=y 6-2 =y x=3 and y=4

Yogita Ingle 2 days, 23 hours ago

2x - y = 2

y = 2x - 2 ...(1)

x + 3y = 15 ...(2)

Substituting the value of y from (1) in (2), we get,

x + 6x - 6 = 15

7x = 21

x = 3

From (1),

y = 2  3 - 2 = 4

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