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gogolik [260]
3 years ago
8

Carl wants to find the product 30×5/6.Which of the following best explains how he can find the product mentally?

Mathematics
1 answer:
prohojiy [21]3 years ago
7 0

Answer:

D

30×5/6 = 25

Step-by-step explanation:

:))

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5/6(z+10)=20 what is z plzz I WILL GIVE YOU 16PTS
aleksandrvk [35]

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1/6

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Determine the missing angle in each triangle.
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45

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3 years ago
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Explain how you would create and use a spinner to simulate the probability of an event with the following probabilities:
castortr0y [4]

To simulate the probability in a spinner divide the spinner into colors depending on the probability each color represents.

<h3>What does the word "probability" mean?</h3>

This word refers to the likelihood for an event to occur. Moreover, it is often expressed either as a fraction, a percentage, or a number.

<h3>How to create a spinner to simulate probability?</h3>

In this case, each of the colors represents a probability:

  • Green 20%
  • Blue 1/4 or 25%
  • Yellow 2/5 oe 40%
  • Orange 15%

Due to this, the best is to divide the spinner by colors and considering the percentanges of each color. For example 40% of the spinner should be red, while 15% should be orange.

Learn more about spinner in: brainly.com/question/24280611

#SPJ1

8 0
2 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 am doing bad in school i dont understand it i am feeling depressed wish someone would just do my work for me sigh
olga2289 [7]

Answer:

help you

Step-by-step explanation:

I will help you on somethings

6 0
3 years ago
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