How do I solve for h?

Mathematics

1 answer:

Darya [45]2 years ago

8 0

Use the Pythagorean theorem.

The answer is 5 $\sqrt{3}$

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(x+5)(x-3)<br> (x-3)(x+10)

antiseptic1488 [7]

I’m learning this right now actually! The first one is...pretend the 2 is small. x2+2-15 and the second answer is x2+7-30!

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3 years ago

At an Italian store the price for T-shirt is 55 euros what is the price in US dollars

Zinaida [17]

Answer:

the current converson rate for euros to dollars is $1 (USD) = 0.91 Euros

so 55 euros would equal about 60.60 USD...

in other words, that better be one nice shirt

Step-by-step explanation:

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Answer:

C. 91.46

Step-by-step explanation:

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ASHA 777 [7]

$\bf tan\left( \frac{x}{2} \right)+\cfrac{1}{tan\left( \frac{x}{2} \right)}\\\\
-----------------------------\\\\
tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\

\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}
\\ \quad \\

\boxed{\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}}
\end{cases}\\\\$

$\bf -----------------------------\\\\
\cfrac{1-cos(x)}{sin(x)}+\cfrac{1}{\frac{1-cos(x)}{sin(x)}}\implies \cfrac{1-cos(x)}{sin(x)}+\cfrac{sin(x)}{1-cos(x)}
\\\\\\
\cfrac{[1-cos(x)]^2+sin^2(x)}{sin(x)[1-cos(x)]}\implies 
\cfrac{1-2cos(x)+\boxed{cos^2(x)+sin^2(x)}}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{1-2cos(x)+\boxed{1}}{sin(x)[1-cos(x)]}\implies \cfrac{2-2cos(x)}{sin(x)[1-cos(x)]}
\\\\\\
\cfrac{2[1-cos(x)]}{sin(x)[1-cos(x)]}\implies \cfrac{2}{sin(x)}\implies 2\cdot \cfrac{1}{sin(x)}\implies 2csc(x)$

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2 years ago

If you are paid $2.20 an hour, worked for 26 hours this

erastova [34]

Answer:

Step-by-step explanation:

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2 years ago

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