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3 years ago

9

Consider the two functions.

1 answer:

Verizon [17]3 years ago

3 0

Sorry, i think you forgot to place the second function. please do so!

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A store discounts a stereo from 130 to 87.50. What is the percent discount?

Minchanka [31]

42.5 percent

Subtract 130 and 87.50

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3 years ago

Find the unit rate of 8 teaspoon for 4 cups.

Igoryamba

Answer:

2 teaspoons for 1 cup

Step-by-step explanation:

8 divided by 4 is 2

4 divided by 4 is 1

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3 years ago

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Find the slope/rate of change of the line on the graph.

ohaa [14]

Answer:

y=-1

From where the line crosses the y axis, it goes up 1 and over 1 from each plane although its moving to the left from 0 every time. This means the slope is negative.

In the equation y=mx+b, b is the y intercept and m is the slope

plugin the info and you get y=-1x+0 or

y=-1x

Step-by-step explanation:

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2 years ago

What is 2^5 in standard notation

elena-14-01-66 [18.8K]

2x2x2x2x2 is 2^5 in standard notation

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2 years ago

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Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

$\frac{\sin \beta}{\cos^2\beta}=\sqrt{2}$

Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

This is a quadratic equation in $\sin \beta$ .

By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

4 0

3 years ago