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3 years ago

What is the solution set of this equation 2|x – 1| – 7 = 2

1 answer:

6 0

$2|x - 1| - 7 = 2\\
2|x-1|=9\\
|x-1|=\dfrac{9}{2}\\
x-1=\dfrac{9}{2} \vee x-1=-\dfrac{9}{2}\\
x=\dfrac{11}{2} \vee x=-\dfrac{7}{2}$

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Stolb23 [73]

Answer:

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4 years ago

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WILL GIVE BRAINLIEST find the value of r (-4,8), (r,12), m=4/3 (show your work pls)

OLEGan [10]

Answer:

r = 13

Step-by-step explanation:

The first step is to find b from y=mx=b using m=4/3 and (-4,8)

$8=\frac{4}{3}(-4)+b$

After solving the equation, b will equal -16/3 making the final equation be

$y=\frac{4}{3} x-\frac{16}{3}$

Using the equation, you can solve for r.

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This will bring the answer being r = 13

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3 years ago

Rationalize the denominator of square root of negative 16 over open parentheses 1 plus i close parentheses plus open parentheses

emmasim [6.3K]

$\bf \cfrac{\sqrt{-16}}{(1+i)+(6+3i)}\implies \cfrac{\sqrt{-1\cdot 16}}{1+i+6+3i}\implies \cfrac{\sqrt{-1}\cdot \sqrt{16}}{7+4i}
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\cfrac{i\cdot \sqrt{4^2}}{7+4i}\implies \cfrac{4i}{7+4i}\impliedby 
\begin{array}{llll}
\textit{now, we'll multiply by the}\\
\textit{conjugate of the denominator}
\end{array}\\\\
-------------------------------\\\\$

$\bf \textit{and recall }\textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
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\\\\\\
\cfrac{16}{65}+\cfrac{28i}{65}$

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4 years ago

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lord [1]

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4 years ago

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skelet666 [1.2K]

Answer:

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3 years ago