Pls help me on this question.Do NOT answer 13 and 15. Only answer 14.
2 answers:
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The only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54
remember if we can do
xy=0 then assume x and y=0
so factor
0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is
if you have
ax^2+bx+c=0 then
x=
x^2+7x+18
a=1
b=7
c=18
x=
x=
x=
i=√-1
x=
the zerose would be
x=
or 
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54
remember if we can do
xy=0 then assume x and y=0
so factor
0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is
if you have
ax^2+bx+c=0 then
x=
x^2+7x+18
a=1
b=7
c=18
x=
x=
x=
i=√-1
x=
the zerose would be
x=
Answer:
256
Step-by-step explanation:
A calculator works well for this.
_____
None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.
1 to any power is still 1, so the factors (1^n) can be ignored.
After you ignore all of the things that can be ignored, your problem simplifies to ...
(2^2)(2^-3)^-2
The rules of exponents applicable to this are ...
(a^b)^c = a^(b·c)
(a^b)(a^c) = a^(b+c)
Then your product simplifies to ...
(2^2)(2^((-3)(-2)) = (2^2)(2^6)
= 2^(2+6)
= 2^8 = 256
