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3 years ago

What is absolute value???

2 answers:

6 0

<span>To remove any negative sign in front of a number, and to think of all numbers as positive (or zero). Example: |5|=5</span>

Ierofanga [76]3 years ago

6 0

Basically I think it of as how far away is this number from 0. So lets say I have a -6 and a positive 6, if they had a Absolute value around them each answer would be 6 due to them both being 6 units away from 0. Its best if you can picture this as a number line.
Hope this helps :D

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What are factors of 27?

Artyom0805 [142]

The answer would be 3^33

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3 years ago

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2 years ago

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almond37 [142]

Answer:

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5 0

2 years ago

Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)

Leni [432]

Alrighty

remember some rules
$(ab)^c=(a^c)(b^c)$
and
$x^{-m}=\frac{1}{x^m}$
and
$x^0=1$ for all real values of x
and
$(a^b)^c=a^{bc}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
(a/b)/(c/d)=(ad)/(bc)
and
$(a^b)(a^c)=a^{b+c}$
and
$\frac{a^m}{a^n}=a^{m-n}$
and don't forget pemdas
example: $-x^m=-1(x^m)$ but $(-x)^m=(-1)^m(x^m)$

so

4.
$(\frac{c^{-2}}{2})^{-2}=$

$\frac{(c^{-2})^{-2}}{2^{-2}}=$

$\frac{c^4}{\frac{1}{2^2}}=$

$\frac{c^4}{\frac{1}{4}}=$

$4c^4$

5.

$\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=$

$\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=$

$\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=$

$\frac{(a^4bc^5)b^3}{(-1)(a^2)}=$

$\frac{-a^4b^4c^5}{a^2}=$

$-a^2b^4c^5$

3 0

2 years ago

What isvequivalent to y=23x−6?

RUDIKE [14]

Answer:

Slope = 46.000/2.000 = 23.000

x-intercept = 6/23 = 0.26087

y-intercept = -6/1 = -6.00000

3 0

3 years ago