Question

How much does 93p2 -- 55p + 4 exceed 13p3-5p2+17p - 90?​

Answer 1

Answer:

13p3 +88p2-36p+4.....

Answer 2

$93p^{2} - 55p +4$ exceeds $-13p^{3} + 98p^{2} - 72p + 94$.

Step-by-step explanation:

We have to subtract $13p^{3} - 5p^{2} + 17p - 90$ from $93p^{2} - 55p + 4$.

$(93p^{2} -55p+4) - (13p^{3} - 5p^{2} + 17p -90)$$=> 93p^{2} - 55p + 4 - 13p^3 + 5p^{2} -17p +90$

Now, collecting like terms,

$=> -13p^{3} + 93p^2 +5p^{2} -55p - 17p +4 +90$

$=> -13p^{3} +(93+5)p^{2}-(55+17)p+94\\=> -13p^{3} + 98p^{2} - 72p+94$

Hence, $93p^{2} - 55p +4$ exceeds $-13p^{3} + 98p^{2} - 72p + 94$.