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3 years ago

Can someone please help me!

Mathematics

1 answer:

Alina [70]3 years ago

8 0

Answer:

Step-by-step explanation:

k(x+6)= 4x²+20

k(10)= k(4+6)= 4*4²+20= 64+20= 84

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A baseball is thrown without any spin. Would you consider the baseball to be in motion?

liq [111]

Answer:

yes

Step-by-step explanation:

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3 years ago

A grid model with 100 squares. 73 squares are shaded. What percent of the model is shaded? What is the equivalent ratio? What is

viva [34]

Answer:

73%

Step-by-step explanation:

First, you'll divide the total squares by 100 to get how much is one percent. Then, you divide the shaded squares by the value of 1%. And you'll get how many percents are shaded.

For equivalent decimal, you divide the number by 100.

Then for equivalent decimal, you get the percent and 100.

$p =100/100=1\\73/1=73\\=0.73\\=\frac{73}{100}$

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3 years ago

My cousin borrowed $18,000 for a new car loan for 6 years. He told me that at the end of the loan, he had paid $25,560 for the i

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3 years ago

Hellllllllllp! Someone help

uysha [10]

Answer:

4. (2, 3)

5. (0, 1)

7. (-1, 2)

Step-by-step explanation:

I hope this helps! Have a nice dayy! :)

3 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