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4 years ago

Solve the proportion. Help ASAP!

Mathematics

1 answer:

gogolik [260]4 years ago

5 0

You need to cross multiply.

(t+7)(t+7)=t^2+7t+7t+49. t^2+14t+49
9*25=225

t^2+14t+49=225
Subtract 225 from both sides

t^2+14t- 176=0

(t-8)(t+22) =0

8 and -22

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The area of the parallelagram is 50

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3 years ago

Math grade 9 , Aor B or C or D fassst plzzz

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3 years ago

One student simplified x5 + x5 to x10 . A second student simplified x5 + x5 to 2x5 . Which student is correct? Explain.

dybincka [34]

An expression is defined as a set of numbers, variables, and mathematical operations. The second student is correct.

<h3>What is an Expression?</h3>

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

Given One student simplified x⁵ + x⁵ to x¹⁰. The second student simplified x⁵+x⁵ to 2x⁵. Since the simplification of (x⁵+x⁵) is equal to 2x⁵. Therefore, the second student is correct.

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2 years ago

If tan theta = startfraction 11 over 60 endfraction, what is the value of cot theta?

Dennis_Churaev [7]

The value of $cot\theta$ when $tan\theta =\frac{11}{60}$ comes to be $\frac{60}{11}$ .

Given that trigonometric ratio:

$tan\theta = \frac{11}{60}$

<h3>What is the tangent of an angle?</h3>

The tangent of an angle is the ratio of the opposite side(to that angle) to the adjacent side(to that angle).

So, for the given problem

Opposite side to $\theta$ = 11

Adjacent side to $\theta$ = 60

So, $cot \theta =\frac{AdjacentSide }{OppositeSide}$

$cot\theta =\frac{60}{11}$

Therefore, the value of $cot\theta$ when $tan\theta =\frac{11}{60}$ comes to be $\frac{60}{11}$ .

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2 years ago

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