2/9×5=10/45
3/5×9=27/45
so 3/5 is bigger then 2/9
5/7×5=25/35
3/5×7=21/35
So 5/7 is larger than 3/5
So from least to greatest the numbers are
2/9 then 3/5 then 5/7
So your answer is A
Answer:
40
Step-by-step explanation:
(2x+1/(2x))^5 *(2x -1/(2x))^5
= ((2x)^2 -1/(2x)^2)^5 (a+b)*(a-b) =a2-b2
= (4x^2-1/4(x)^2)^5
now
x =4x^2. ,a = 1/4(x)^2 ,n =5
we have
general term = Cr *x^r *a^(n-r)
= Cr * (4x^2)^r * (1/4(x)^2)^(n-r)
= Cr *4^r * X^2r * 1/( 4^(n-r) *x^(2n-2r)
= Cr * 4^r/4^(n-r) * x^(2r)/x^(2n-2r)
= Cr * 4(2r-n) *x(4r-2n)
now for x^2
4r-2n = 2
4r -10=2
4r =12
r = 3
now for coeff
C(5,3) * 4^(2*3-5)
5!/(3!*(5-3)!) * 4
5*4/(2*1)*4
40
Answer:
1,2,3,2,2
Step-by-step explanation:
because standard form is ax
<h2>~<u>Solution</u> :-</h2>
Here, it is given that the bag contains 25 paise coins and 50 paise coins in which, 25 paise coins are 6 times than that of 50 paise coins. Also, the total money in the bag is Rs. 6.
- Hence, we can see that, here, we have been given the linear equation be;
Let the number of coins of 50 paise will be $ x $ and the number of coins of 25 paise will be $ 6x $ as given. . .
Hence,




- Hence, the number of 50 paise coins will be <u>2</u>. And, 6 times of two be;

- Hence, the number of 25 paise coins will be <u>12</u>.