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3 years ago

Solve each division problem, and classify it based on its quotient.

Mathematics

1 answer:

suter [353]3 years ago

5 0

Answer:

show the image so i can answer

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The following is the formula to find the area of a circle. Which shows the formula correctly solved for the radius (r) in terms

Otrada [13]

Answer:

Step-by-step explanation:

This has to be C because you need to divide pi from r squared bc they r multiplied. Then you get a/pi = r squared. You need to find the square root of a/pi. So r equals square root a/pi.

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3 years ago

Some one please help me ms derienzo is killing my little pea brain

Digiron [165]

I’m pretty sure the answer would be A, sorry if i’m incorrect, have a great day :)

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3 years ago

Mandy is thinking of a prime number that is greater than 41 and less than 47 .what is mandys number.

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

the only prime number in between those numbers is 43.

3 0

3 years ago

Read 2 more answers

6 math geniuses help please!! math

damaskus [11]

Answer:

Step-by-step explanation:

Begin with Euler's formula. I am assuming youre aware of this if youre taking complex algebra? You can prove Euler's formula by doing a Maclaurin expansion of cosine, sine, and e^x. Euler's formula states that:

$e^{ix}=cos(x)+isin(x)$

We can put the first complex number in exponetial form by noticing the input is 2pi/3. The second complex number has an input of pi/3. Therefore:

$z_1=8e^{i(2\pi /3)}$

$z_2=0.5e^{i(\pi /3)}$

Then:

$\frac{z_1}{z_2} =\frac{8e^{i(2\pi /3)}}{0.5e^{i(\pi /3)}}$

Simplify the coefficients to get:

$\frac{z_1}{z_2} =\frac{16e^{i(2\pi /3)}}{e^{i(\pi /3)}}$

When you divide exponentials, you subtract the exponents. Therefore:

$\frac{z_1}{z_2} =16e^{i(2\pi /3)-i(\pi /3)}=16e^{i(\pi /3)$

Put it back into trigonemtric form using Euler's formula:

$16e^{i(\pi /3)}=16cos(\pi /3)+i16sin(\pi /3)$

Cosine of pi/3 is 0.5, and sine of pi/3 is square root of 3 over 2. We have:

$16e^{i(\pi /3)}=16cos(\pi /3)+i16sin(\pi /3)=16*0.5+16*\frac{\sqrt{3} }{2} *i=8+8\sqrt{3} i$

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2 years ago

Let g(x)=2x and h(x)=x^2+4 (g o h)(0)

sweet [91]

Answer would be 0 because whatever you get when you plug in g&h in the equation you have to multiply that by 0. And a number multiplied by 0 is always 0.
Hope this helped. Have a great day!

4 0

4 years ago