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larisa [96]
3 years ago
9

G(x) = 2x + 2 f(x) = 3x - 4 Find (g • f)(-6) -34 -42 -36 38

Mathematics
1 answer:
Kazeer [188]3 years ago
8 0

Answer: f(x)= 3x + 10 and g(x)= 2x - 4

(f+g)(x)= 3x + 10 + 2x - 4

3x + 2x + 6

5x + 6

b and c

Step-by-step explanation:

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3 0
3 years ago
How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
-7x - 12 = - 7х + 3(-4 - x)
dsp73

Answer:

x=0

Step-by-step explanation:

-7x - 12 = - 7x + 3(-4 - x)

Open up the parentheses

-7x - 12 = - 7x -12-3x

-7x cancels out on both sides

-12 =-12-3x

-12 cancels out on both sides

0=-3x\\x=0

I hope this helps!

4 0
3 years ago
What is 7 1 / 5 - 6 2/5
Lunna [17]
7\dfrac{1}{5}=\dfrac{7\cdot5+1}{5}=\dfrac{35+1}{5}=\dfrac{36}{5}\\\\6\dfrac{2}{5}=\dfrac{6\cdot5+2}{5}=\dfrac{30+2}{5}=\dfrac{32}{5}\\\\7\dfrac{1}{5}-6\dfrac{2}{5}=\dfrac{36}{5}-\dfrac{32}{5}=\dfrac{36-32}{5}=\boxed{\dfrac{4}{5}}



7\dfrac{1}{5}-6\dfrac{2}{5}=(7-6)+\left(\dfrac{1}{5}-\dfrac{2}{5}\right)=1+\dfrac{1-2}{5}=1+\dfrac{-1}{5}=1-\dfrac{1}{5}\\\\=\dfrac{5}{5}-\dfrac{1}{5}=\dfrac{5-1}{5}=\boxed{\frac{4}{5}}
4 0
3 years ago
Suppose line n has a slope of 5/7 and passes through (4,8). what is the equation for n in point-slope form?
Olegator [25]

Answer:

\large\boxed{y-8=\dfrac{5}{7}(x-4)}

Step-by-step explanation:

The point-slope form of an equation of a line:

y-y_1=m(x-x_1)

m - slope

We have the slope m=\dfrac{5}{7} and the point (4,\ 8).

Substitute:

y-8=\dfrac{5}{7}(x-4)

7 0
3 years ago
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