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3 years ago

Find the sides and angles please help me

Mathematics

1 answer:

marshall27 [118]3 years ago

3 0

Answer: if this uses the Pythagorean theorem then it’s the square root of 83.24

Step-by-step explanation: the formula is a^2+b^2=c^2 the hypotenuse is c^2

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How to solve 1.25 x 10^-4

enyata [817]

Answer:

0.000125x

Step-by-step explanation:

4 0

3 years ago

A piece of work can be finished in 8 days, what part of the work is done in 2 days

WINSTONCH [101]

Answer:

1/4 of the work

Step-by-step explanation:

A quarter (1/4) of the work is done in 2 days. Let's break it down:

Work can be finished in 8 days total so with 2 days it is only partially completed. If we set it up as 2/8 we can simplify to get 1/4. Therefore, we know only a quarter of the work is done so far.

7 0

3 years ago

Help me answer this question

balandron [24]

Answer: Sorry i whould help but I cant see it can you put it more bigger

Step-by-step explanation:

5 0

3 years ago

I need help please! If you do know the answer help me understand

koban [17]

Answer:

1017.4 m³

Step-by-step explanation:

Volume of cone V = (1/3)πr²h

pi = 3.14

r = 9 and r² = 81 m²

h = 12 m

V = (1/3)*3.14*81*12

V = 1017.36 rounded 1017.4 m³, which is the first answer.

5 0

3 years ago

I need help plz. Its DeMoivres Theorem

vagabundo [1.1K]

$\bf \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ z=\stackrel{a}{-1}\stackrel{b}{-\sqrt{3}~i}~~ \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right)\\[-0.5em] \hrulefill\\ r=\sqrt{(-1)^2+(\sqrt{3})^2}\\ \qquad \sqrt{4}\\ \qquad 2\\ \theta =tan^{-1}\left( \frac{-\sqrt{3}}{-1} \right)\\\\ \qquad \frac{4\pi }{3} \end{cases}$

$\bf z=2\left[ cos\left( \frac{4\pi }{3} \right) -i~sin\left( \frac{4\pi }{3} \right)\right]\implies z^6=\left[ 2\left[ cos\left( \frac{4\pi }{3} \right) -i~sin\left( \frac{4\pi }{3} \right)\right] \right]^6 \\\\\\ z^6=2^6\left[ cos\left( 6\cdot \frac{4\pi }{3} \right) -i~sin\left( 6\cdot \frac{4\pi }{3} \right)\right]\implies z^6=64[cos(8\pi )-i~sin(8\pi )] \\\\\\ z^6=64[(-1)-i~(0)]\implies z^6=\stackrel{\stackrel{a}{\downarrow }}{-64}~~\stackrel{\stackrel{b}{\downarrow }}{-0i}$

4 0

4 years ago