I believe it would be 6. the first number and exponent are usually your quadratic terms
The answer is 2.
p=3
3^2+5(4 x 3)= 2
We could do it with algebra. But we can also do it the long way, which is
shorter than the algebraic way.
The digit in the tens place is 3 times the digit in the units place.
So the number MUST be
31, or
62, or
93 .
It can't be anything else.
Now here they are again, with the reverse of each one:
31 . . . 13 The new number is 18 less.
62 . . . 26 The new number is 36 less.
93 . . . 39 The new number is 54 less.
Obviously, the original number is 62.
Answer:
A. 4(x - 4) + 2(3x² + 3x - 20)
C. (11x² + 7x - 55)-(5x² - 3x + 1)
F. (3x² + 5x - 28) + (3x² + 5x - 28)
Step-by-step explanation:
Given:
(3x-7)(2x+8)
= 6x² + 24x - 14x - 56
=6x² + 10x - 56
A. 4(x - 4) + 2(3x² + 3x - 20)
= 4x - 16 + 6x² + 6x - 40
= 6x² + 10x - 56
B. (3x² + 5x - 28) - (2x² + 4x + 28)
= 3x² + 5x - 28 - 2x² - 4x - 28
= x² + x - 56
C. (11x² + 7x - 55)-(5x² - 3x + 1)
= 11x² + 7x - 55 - 5x² + 3x - 1
= 6x² + 10x - 56
D. 4(x - 4) - 2(3x² + 3x - 20)
= 4x - 16 - 6x² - 6x + 40
= - 6x² - 2x + 24
E. (11x² + 7x - 55)-(5x² - 3x + 2)
= 11x² + 7x - 55 - 5x² + 3x - 2
= 11x² - 5x² + 7x + 3x - 55 - 2
= 6x² + 10x - 57
F. (3x² + 5x - 28) + (3x² + 5x - 28)
= 3x² + 5x - 28 + 3x² + 5x - 28
= 6x² + 10x - 56
