Question

Subtract 11b^2-4b+711b 2 −4b+711, b, start superscript, 2, end superscript, minus, 4, b, plus, 7 from b^2+8b-9b 2 +8b−9b, start superscript, 2, end superscript, plus, 8, b, minus, 9.

Answer 1

Answer:-730b^2-1404b-16

Step-by-step explanation:

11b^2-4b+711b2-4b+711b^2-4b+7=

(11+711)b^2+(-4+711.2-4-4)b+7= 722b^2+1410b+7 =A

b^2+8b-9b2+8b-9b^2+8b-9=

(1-9)b^2+(8-9.2+8+8)b-9= -8b^2+6b-9 =B

B-A= (-8-722)b^2+(6-1410)b+(-9-7)= -730b^2-1404b-16

Answer 2

Answer:

$-10b^2+12b-16$

Step-by-step explanation:

Consider the given question is " Subtract $11b^2-4b+7$ from $b^2+8b-9$".

$(b^2+8b-9)-(11b^2-4b+7)$

Using distributive property we get

$b^2+8b-9-(11b^2)-(-4b)-(7)$

$b^2+8b-9-11b^2+4b-7$

On combining like terms we get

$(b^2-11b^2)+(8b+4b)+(-9-7)$

$-10b^2+12b+(-16)$

$-10b^2+12b-16$

Therefore, the resultant expression is $-10b^2+12b-16$.