Factor:
3x^2 + 27
= 3(x^2 + 9)
Answer is 3(x^2 + 9), when factored.
A) (3x + 9i)(x + 3i)
= (3x + 9i)(x + 3i)
= (3x)(x) + (3x)(3i) + (9i)(x) + (9i)(3i)
= 3x^2 + 9ix + 9ix + 27i^2
= 27i^2 + 18ix + 3x^2
B) (3x - 9i)(x + 3i)
= (3x + - 9i)(x + 3i)
= (3x)(x) + (3x)(3i) + ( - 9i)(x) + (- 9i)(3i)
= 3x^2 + 9ix - 9ix - 27i^2
= 27i^2 + 3x^2
C) (3x - 6i)(x + 21i)
= (3x + - 6i)(x + 21i)
= (3x)(x) + (3x)(21i) + (- 6i)(x) + ( -6i)(21i)
= 3x^2 + 63ix - 6ix - 126i^2
= - 126i^2 + 57ix + 3x^2
D) (3x - 9i)(x - 3i)
= (3x + - 9)(x + - 3)
= (3x)(x) + (3x)( - 3i) + (- 9)(x) + ( - 9)( - 3i)
= 3x^2 - 9ix - 9x + 27i
= 9ix + 3x^2 + 27i - 9x
Hope that helps!!!
2,320,502 <=== Two million, three-hundred and twenty thousand five hundred and two
B is your answer.
Because 7 times 8 is 56 and 7 times 3 is 21.
Answer:
see below
Step-by-step explanation:
|x - 3| < 4
To remove the absolute values, we need to make two inequalities, one positive and one negative
x-3 < 4 and x-3 > -4
Add 3 to each side
x-3+3<4+3 and x-3+3 > -4+3
x<7 and x >-1
-1 <x <7
TWO SCREENSHOT BELOW FOR THE ANSWER
