What is th product ? y2+3y+7 8y2+y+1

8y⁴+25y³+60y²+10y+7

Step-by-step explanation:

Given two functions

y²+3y+7 and 8y²+y+1

To take the product of both functions means multiplying both functions together.

(y²+3y+7)(8y²+y+1)

Opening up the bracket we have:

= 8y⁴+y³+y²+24y³+3y²+3y+56y²+7y+7

Collecting the like terms

= 8y⁴+y³+24y³+y²+3y²+56y²+3y+7y+7

Adding up the functions with the same degree and simplifying the expression will give;

= 8y⁴+25y³+60y²+10y+7

This gives the product of both functions

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.