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mixer [17]
3 years ago
13

NEED DONE ASAP. IT'S 12AM AND I WANNA GET THIS DONE SO I CAN SLEEP..

Mathematics
1 answer:
mr Goodwill [35]3 years ago
8 0

The options seem a bit mis-formatted but here is the way p is computed:

the decrease is 2000-1600

and it is relative to: 2000  (2003 attendance)

so p=100 \frac{2000-1600}{2000} = 100 \frac{400}{2000}

and so that aligns with your Option A. Options B, C, D look incorrect for sure.

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Noelle has set a goal to run more than 50 mi this month. So far, she has run 10 mi. Her average running speed is 5 mph.
Vladimir79 [104]

Answer:

To meet her goal, Noelle must run more than 8\ hours the rest of the month

Step-by-step explanation:

Let

h-------> the number of hours Noelle runs for the rest of the month

we know that

The inequality that represent the situation is

10+5h>50

Solve for h

Subtract  10 both sides

5h> 50-10

5h> 40

Divide by 5 both sides

h> 8\ hours

3 0
3 years ago
What is the measure of ABD?
Mrac [35]

Answer:

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3 0
3 years ago
Read 2 more answers
Can someone help me with this question please. I'll give the brainliest answer to whoever helps me.
Doss [256]

Answer: \sqrt[5]{y}

I realize its probably not the largest readable font. If you are having trouble reading it, it is the square root of y; however, there is a tiny little 5 in the upper left corner to indicate a fifth root. So you would read it out as "the fifth root of y"

The rule I'm using is

x^{1/n} = \sqrt[n]{x}

and the more general rule we could use is

x^{m/n} = \sqrt[n]{x^m}

where m = 1. This rule helps convert from rational exponent form (aka fractional exponents) to radical form.

5 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
ASAP!!!!!!!<br><br> y-13=4(x-2)
Juliette [100K]
Y-13=4(x-2)
y-13=4x-8
y=4x+5
5 0
3 years ago
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