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lara [203]
3 years ago
12

Please answer! Thanks!

Mathematics
1 answer:
padilas [110]3 years ago
8 0
100 - 6 (3)
100 - 6 (7)
100 - 6 (14)
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Divide el rufini6x-4x+2 entre x-2<br> x+2x+x+2 entrex+2
Mashcka [7]

Answer:

Combine Like terms and do the Equation normally

Step-by-step explanation:

8 0
3 years ago
(b)The graph of y = g(x) is shown. Draw the graph of g(2x). Will award 100 Brainly points
ivanzaharov [21]

Using the stated transformation, the graph of g(2x) is given at the end of the answer.

<h3>Horizontal stretch and compression</h3>

An horizontal stretch or an horizontal compression happens when a constant is multiplied at the domain of the function, as follows:

g(x) = f(ax).

The definition of stretch or compression depends on the value of the constant a, as follows:

  • If a > 1, it is a compression by a factor of 1/a.
  • If a < 1, it is a stretch by a factor of 1/a.

In this problem, the rule is:

f(x) = g(2x).

Meaning that f(x) is an horizontal compression by a factor of 1/2 of g(x), and then the vertices are given as follows:

  • (0,0).
  • (1,-4).
  • (2,-2).

That is, in each vertex, the x-coordinate was divided by 2, and thus the graph with these vertices is given at the end of the answer.

More can be learned about transformations at brainly.com/question/28725644
#SPJ1

8 0
1 year ago
Add 8x to 2x and then subtract 5 from the sum. If x is a
Anuta_ua [19.1K]

Answer:10

Step-by-step explanation:

7 0
3 years ago
What is the distance between (5,2) and (1,-3) on a coordinate plane?
Citrus2011 [14]
I believe it would be (4,5)
6 0
3 years ago
Read 2 more answers
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
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