Question

5(3y-2)(3y+5)=Ay²+By+C is true'for all value of y calculate A-B+C​

Answer 1

Step-by-step explanation:

5(3y-2)(3y+5)=Ay²+By+C

(15y-10)(3y+5)=Ay²+By+C

45y²+75y-30y-50=Ay²+By+C

45y² +45y-50 =Ay²+By+C

A = 45,

B = 45

C = -50

A-B+C= 45- 45 +(-50)

= 45+45-50

= -50

Answer 2

The correct answer is -50.

Given: The equation = 5(3y-2)(3y+5) = Ay²+By+C.

To Find: The value of A - B + C.

Solution:

5(3y-2)(3y+5)

= $5(9y^{2} -6y+15y-10)$

= $5(9y^{2} +9y-10)$

= $45y^{2} +45y-50$

Now compare the equation $45y^{2} +45y-50$ and Ay²+By+C.

Coefficient of $y^{2}=A = 45$.

Coefficient of y = B = 45.

Coefficient = constant = C = -50.

So, now in A-B+C put the values of A, B and C.

A-B+C = 45 - 45 + (-50)

A-B+C = -50

Hence, the value of A-B+C is -50.

#SPJ5