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cluponka [151]
3 years ago
12

If one supplementary angle measure 51 degrees the other angle measures 39 degrees?

Mathematics
1 answer:
mojhsa [17]3 years ago
7 0
The answer to your question is false
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I Need Help With Statistics
Ilya [14]

Answer: Choice A) mean, there are no outliers

Have a look at the image attached below. I made two dotplots for the data points. The blue points represent bakery A. The red points represent bakery B. For any bakery, the points are fairly close together. There is no point that is off on its own. So there are no outliers, making the mean a good choice for the center. If there were outliers, then the median is a better choice. The mean is greatly affected by outliers, while the median is not.

5 0
3 years ago
Convert 12.3 liters to milliliters
kifflom [539]

Answer:

12300 militers

Step-by-step explanation:

1 milileter = .001 liter

12.3 liters = 100 x 12.3

7 0
3 years ago
5. Solve for x using any method you'd like. Round to the hundredth. Show all your work. (2 points each) d. 20x^3 = 7x^5
Sindrei [870]

Solving this equation for x, we have:

\begin{gathered} 20x^3=7x^5 \\ 7x^5-20x^3=0 \\ x^3(7x^2-20)=0 \\ \begin{cases}x^3=0\to x=0 \\ 7x^2=20\to x=\pm\sqrt[]{\frac{20}{7}}\end{cases} \end{gathered}

6 0
11 months ago
How many grams are in 15 kilograms? 1.5 150 1,500 15,000
diamong [38]
15000 because 15kg x 1000 = 15000g
8 0
3 years ago
Read 2 more answers
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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