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kherson [118]
1 year ago
6

A support wire reaches from the top of a pole to a clamp on the ground. The pole is

Mathematics
1 answer:
Burka [1]1 year ago
6 0

The length of the support wire on the pole is 27 feet

<h3>How to determine the length of the support?</h3>

The given parameters are:

  • Distance from the base, b = 10 feet
  • Angle with the ground, x = 68 degrees

The length (L) of the support wire is then calculated using the following cosine function

cos(x) = b/L

This gives

cos(68) = 10/L

Make L the subject

L = 10/cos(68)

Evaluate

L = 27

Hence, the length of the support wire is 27 feet

Read more about triangles at:

brainly.com/question/1675117

#SPJ1

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Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of z=0-2i. First, convert from Cartesian to polar form:

r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2

\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}

z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})

Next, use the formula \displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr] where \displaystyle k=0,1,2,...\:,n-1 to find the square roots:

<u>When k=1</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]

\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)

\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i

<u>When k=0</u>

<u />\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]

\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)

\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i

Thus, the square roots of -2i are 1-i and -1+i

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