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xxTIMURxx [149]
3 years ago
11

Help me ASPA PLEASE help

Mathematics
1 answer:
Olegator [25]3 years ago
4 0

Answer:

85

Step-by-step explanation:

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15 POINTS!! PLEASE ANSWER AS FAST AS POSSIBLE!!
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Answer:

-13

Step-by-step explanation:

-5-8= -13

6 0
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Given f(x) = 2x and h(x) = 2 - 6x<br>Find (f + h) (x)<br><br><br>​
gulaghasi [49]

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idk

Step-by-step explanation:

6 0
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Which expression represents "6 more than x"?
zysi [14]

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Step-by-step explanation:

6 0
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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
3 years ago
You are given f(0) = 1 and g(0) = -2 and g(1) = 3. What is <br> g ∘ f ( 0 ) <br> equal to?
Sergeu [11.5K]

Answer:

  • <u><em>g∘f (0) = 1</em></u>

Explanation:

The <em>composition</em> of the <em>functions</em> f and g represented by g ∘ f ( 0 ) means that g is applied to f(0), i.e f(0) is the input to the function g.

Since f(0) = 1, you are going fo find g(1):

         \( g\circ f(x)=g(f(x)) \)\\\\\( g\circ f(0)=g(f(0)) \)\\\\\( g\circ f(0)=g(1) \)=3

4 0
3 years ago
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