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il63 [147K]
3 years ago
5

4. An extension ladder is leaned up against a building to reach a window that is 16.1 feet

Mathematics
1 answer:
ElenaW [278]3 years ago
3 0

Answer:

a) ~18.95ft

b) ~58°

c) ~32°

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What is the slope of the line that passes through (-2, 7) and (4, 9)
telo118 [61]

Answer: The slope is 1/3

Step-by-step explanation:

m = y2-y1/x2-x1

m =9-7/4+2

m=2/6

m=1/3

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2 years ago
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You are debating whether to buy a new computer for $1,360.00 with a discount of
olasank [31]

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$167.03

Step-by-step explanation:

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Paul spent 3/8 of her allowance on clothes in 16 on Entertainement what fraction of her allowance did she spend on other items
nasty-shy [4]

Answer:

11 / 24

Step-by-step explanation:

i'm assuming you mean 1/6


We have to calculate the fraction of Paula`s allowance that she spent on other items if she already had spent 3/8 on clothes and 1/6 on entertainment. First we have to add: 3 / 8 + 1 / 6 = ( LCD is 24 ) = 9 / 24 + 4 / 24 = 13 / 24. Then : 1 - 13 / 24 = 24 / 24 - 13 / 24 = 11 / 24. Answer: She has spent 11 / 24 of her allowance on other items.



6 0
3 years ago
4ab (a -b) + b (2b -2a²) for a = 1 and b= -1<br> simplify it
aleksley [76]

Answer:

Step-by-step explanation:

4ab(a-b)+b(2b-2a^2)

=4*1*-1[1-(-1)]+(-1){2(-1)-2*1^2}

=-4(1+1)-1{-2-2}

=-4*2-1(-4)

=-8+4

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3 0
3 years ago
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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
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