Question

By how much does 3xcube - 5xsquare plus 2x-3 exceed 2xcube - 3xsquare plus x plus 1
Plz help with working

Answer 1

Answer:

(i) b × b × b × ... 15 times

(ii) y × y × y × ... 20 times

(iii) 14 × a × a × a × a × b × b × b

(iv) 6 × x × x × y × y

(v) 3 × z × z × z × y × y × x

Answer 2

Answer:

2x+5

Step-by-step explanation:

since the question how much does  3xcube - 5xsquare plus 2x-3 exceed 2xcube - 3xsquare plus x plus 1, we will use subtraction.

3xcube  - 5xsquare + 2x-3 - 2xcube - 3xsquare + x +1

(3x cube = 27, 5x square therefore 25-27+2 = x)

x-3-2xcube - 3xsquare + X + 1

(2xcube = 8 3xsquare =9 therefore 8-9 = -1, and since 3 and 1 are normal numerals we'll add them = +1+3=4)

x- (-1) +X+4

(now we add the the X+X and the numerals 4+1)

Ans: 2x+5

Hope It Helps You!!