Question

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

Answer 1

to find,

13p^2-5p^2+17p-90+x=93p^2-55p^2+4

we need to find x in the above equation

8p^2+17p-90+x=38p^2+4

x=38p^2+4-(8p^2+17p-90)

x=38p^2+4-8p^2-17p+90)

x=30p^2-17p+94

hence shown

Answer 2

Answer:

Step-by-step explanation:

solution:
Step 1:In order to get the result, we 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

Step 2:Combine the like terms and simplify

​

 

=−13p

3

+(93p

2

+5p

2

)+(−55p−17p)+(4+90)

=>93p

2

−55p+4−(13p

3

−5p

2

+17p−90)=−13p

3

+98p

2

−72p+94

​

Hence, 93p

2

−55p+4 exceeds 13p

3

−5p

2

+17p−90 by −13p

3

+98p

2

−72p+94