# Answer on Algebra Question for kelsi

Question #5522

simplify each expression then evaluate the expressionfor the given value of the variable

-b^2(2b+4)+b^5,b=-1

3c^2+1+(5c)^2,c=3

20-2x^2/x,x=-2

y^-3(2y/9),y=-3

-b^2(2b+4)+b^5,b=-1

3c^2+1+(5c)^2,c=3

20-2x^2/x,x=-2

y^-3(2y/9),y=-3

Expert's answer

simplify each expression then evaluate the expression for the given value of the variable

1)& -b^2(2b+4)+b^5, b=-1

-b^2(2b+4)+b^5 = -(-1)^2(2(-1)+4)+(-1)^5 = -1(-2+4)+(-1) = -1(-2+4)+(-1) = -2-1 = -3.

2)& 3c^2+1+(5c)^2,c=3

3c^2+1+(5c)^2 = 3*3^2+1+(5*3)^2 = 3^3+1+15^2 = 27+1+225 = 253.

3) 20-2x^2/x,x=-2

20-2x^2/x = 20-2(-2)^2/(-2) = 20-2*4/(-2) = 20+2*4/2 = 20+4 = 24.

4)y^(-3)(2y/9),y=-3

y^(-3)(2y/9) = (-3)^(-3)(2(-3)/9) = (-3)^(-3)(-6/9) = (-1/3^3)(-6/9) = (1/3^3)(6/9) = (1/27)(6/9) = 6/(9*27) = 2/(3*27) = 2/81.

1)& -b^2(2b+4)+b^5, b=-1

-b^2(2b+4)+b^5 = -(-1)^2(2(-1)+4)+(-1)^5 = -1(-2+4)+(-1) = -1(-2+4)+(-1) = -2-1 = -3.

2)& 3c^2+1+(5c)^2,c=3

3c^2+1+(5c)^2 = 3*3^2+1+(5*3)^2 = 3^3+1+15^2 = 27+1+225 = 253.

3) 20-2x^2/x,x=-2

20-2x^2/x = 20-2(-2)^2/(-2) = 20-2*4/(-2) = 20+2*4/2 = 20+4 = 24.

4)y^(-3)(2y/9),y=-3

y^(-3)(2y/9) = (-3)^(-3)(2(-3)/9) = (-3)^(-3)(-6/9) = (-1/3^3)(-6/9) = (1/3^3)(6/9) = (1/27)(6/9) = 6/(9*27) = 2/(3*27) = 2/81.

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment