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SVETLANKA909090 [29]
4 years ago
11

Which expressions are equivalent to RootIndex 3 StartRoot 128 EndRoot Superscript x? Select three correct answers.

Mathematics
1 answer:
Virty [35]4 years ago
7 0

Answer:

(A)128^{x/3}

(D)(4(2^{1/3}))^x

Step-by-step explanation:

We want to determine which of the expression is equivalent to:

\sqrt[3]{128}^ x

By the law of indices:

\sqrt[3]{128}=128^{1/3}\\$Therefore:\\\sqrt[3]{128}^ x \\=(128^{1/3})^x\\=128^{x/3}

Similarly:

\sqrt[3]{128}^ x \\=\sqrt[3]{64*2}^ x\\=(4\sqrt[3]{2})^ x\\=(4(2^{1/3}))^x

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line m contains the points -3,4 and 1,0. write the equation of a line that would be perpendicular to this one and pass through t
castortr0y [4]

The equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6) in point slope form is y = x + 8

<h3><u>Solution:</u></h3>

Given that line m contains points (-3, 4) and (1, 0)

We are asked to find the equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6)

<em><u>Let us first find slope of the line "m"</u></em>

Given two points are (-3, 4) and (1, 0)

m = \frac{y_2 - y_1}{x_2 - x_1}

\left(x_{1}, y_{1}\right)=(-3,4) \text { and }\left(x_{2}, y_{2}\right)=(1,0)

m=\frac{0-4}{1-(-3)}=-1

Thus slope of line m is -1

We know that <em>product of slope of given line and perpendicular line are always -1</em>

So, we get

\begin{array}{l}{\text { slope of line } m \times \text { slope of perpendicular line }=-1} \\\\ {-1 \times \text { slope of perpendicular line }=-1} \\\\ {\text { slope of perpendicular line }=1}\end{array}

So we have got the slope of perpendicular line is 1 and it passes through (-2, 6)

Let us use the point slope form to find the required equation

<em><u>The point slope form is given as:</u></em>

y - y_1 = m(x - x_1)

(x_1, y_1) = (-2, 6) and m = 1

y - 6 = 1(x - (-2))

y - 6 = x + 2

y = x + 8

Thus equation of required line in point slope form is y = x + 8

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