Find the area of this parallelogram.

Mathematics

1 answer:

Alex17521 [72]3 years ago

8 0

Answer:

it will be c

Step-by-step explanation:

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√(11 - 6 √(2))<br> (not decimal form)

bogdanovich [222]

Answer:

$\huge\boxed{\sqrt{11-6\sqrt2}=3-\sqrt2}$

Step-by-step explanation:

$\sqrt{11-6\sqrt2}=(*)\\\\11=9+2=3^2+(\sqrt2)^2\\6\sqrt2=(2)(3)(\sqrt2)\\\\(*)=\sqrt{3^2-(2)(3)(\sqrt2)+(\sqrt2)^2}\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=\sqrt{(3-\sqrt2)^2}\\\\\text{use}\ \sqrt{a^2}=|a|\\\\=|3-\sqrt2|=3-\sqrt2\\\\^{\text{because}\ 3>\sqrt2\to3-\sqrt2>0}$

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3 years ago

What is tje solution to the inequality 6x-5>-29

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Step-by-step explanation:

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For his long distance phone service, Chris pays a $2 monthly fee plus 12 cents per minute. Last month, Chris's long distance bil

photoshop1234 [79]

Use the terms and variables given to find an equation to find the answer.
Price= $2 + .12xminutes
P=2+0.12m

Then, plug in known variables to find the final answer.
15.08=2+0.12m
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4 years ago

Means to find the only possible answer.

erastova [34]

The answer is determine

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3 years ago

If z1= 3+3i and z2=7(cos(5pi/9) + i sin (5pi/9)), then z1/z2= blank

mixas84 [53]

$z1=\stackrel{a}{3}+\stackrel{b}{3}i~~ \begin{cases} r = \sqrt{a^2+b^2}\\ r = \sqrt{18}\\[-0.5em] \hrulefill\\ \theta =\tan^{-1}\left( \frac{b}{a} \right)\\ \theta =\frac{\pi }{4} \end{cases}~\hfill z1=\sqrt{18}\left[\cos\left( \frac{\pi }{4} \right) i\sin\left( \frac{\pi }{4} \right) \right] \\\\[-0.35em] ~\dotfill$

$\cfrac{z1}{z2}\implies \cfrac{\sqrt{18}\left[\cos\left( \frac{\pi }{4} \right) i\sin\left( \frac{\pi }{4} \right) \right]} {7\left[\cos\left( \frac{5\pi }{9} \right) i\sin\left( \frac{5\pi }{9} \right) \right]} \\\\[-0.35em] ~\dotfill\\\\ \qquad \textit{division of two complex numbers} \\\\ \cfrac{r_1[\cos(\alpha)+i\sin(\alpha)]}{r_2[\cos(\beta)+i\sin(\beta)]}\implies \cfrac{r_1}{r_2}[\cos(\alpha - \beta)+i\sin(\alpha - \beta)] \\\\[-0.35em] ~\dotfill$

$\cfrac{z1}{z2}\implies \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{\pi }{4}-\frac{5\pi }{9} \right)+i\sin\left( \frac{\pi }{4}-\frac{5\pi }{9} \right) \right] \\\\\\ \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{-11\pi }{36} \right) +i\sin\left( \frac{-11\pi }{36} \right) \right]\implies \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{83\pi }{36} \right) +i\sin\left( \frac{83\pi }{36} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{z1}{z2}\approx 0.348~~ + ~~0.496i~\hfill$

6 0

3 years ago