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steposvetlana [31]

2 years ago

6

Ms. Fitzsimmons wanted to buy two of the same jackets in different colors online. The online store

2 answers:

zepelin [54]2 years ago

4 0

Answer:

the cost of one jacket is $27.14

Sonbull [250]2 years ago

4 0

Shejdirkendbehruejenrbrififjfnjdjeeiwiwiwjnsbddb

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Is -128/5 a terminating or repeating decimal??

hjlf

Answer:

terminating

Step-by-step explanation:

-128 / 5 = -25.6

-25/6 is terminating because it has an end. a repeating decimal on the other hand goes on forever, like 3.33333...

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3 years ago

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Black_prince [1.1K]

The answer is attached.

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3 years ago

Find the value of − √49

melomori [17]

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2 years ago

What is 9.375 Rounded to the nearest hundredth

stiks02 [169]

Note the hundredth place value (underlined and bolded):

9.3<u>7</u>5

Look at the number to the right of the hundredth place value. It is a 5. Because you round up if the number is 5 or greater, you round up in this case (you round down if it is 4 or less).

9.375 rounded to the nearest hundredth is 9.38

~

3 0

3 years ago

Read 2 more answers

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