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3 years ago

Find the function value. cos150°

Mathematics

1 answer:

Paladinen [302]3 years ago

7 0

Answer:

$cos150 = -\frac{\sqrt{3} }{2}$

Step-by-step explanation:

Recall the unit circle. At 150 deg, the point value is (-sqrt3/2, 1/2)

Remember the cosine is always the x-value, and sine is always the y-value.

This means that cosine will be -sqrt3/2.

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Of the words, which is five-sixths of the list. How many words are on the list?

BlackZzzverrR [31]

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3 years ago

5a + 2b is a A) monomial B) binomial C) trinomial

TiliK225 [7]

Answer:

binomial

Step-by-step explanation:

A binomial is a polynomial with 2 terms

5a+2b has two terms

8 0

3 years ago

Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago

I need number 4 solved please

valentina_108 [34]

Answer:

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Step-by-step explanation:

7 0

3 years ago

simple random sample from a population with a normal distribution of 99 body temperatures has x overbarequals98.20degrees Upper

Verizon [17]

Answer:

Step-by-step explanation:

Let x be the normal body temperature

X is N(99, sigma)

Sample mean x bar = 98.2 and s = 0.63

Margin of error = Z critical * 0.63 = 1.95 *0.63 = 1.2285

It is safe since 1.5 <1.96 critical value of z at 95%

7 0

3 years ago