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3 years ago

15

Find the Area of the figure below, composed of a rectangle and a semicircle. The radius of the circle is shown. Round to the nea

rest tenths place.

1 answer:

liraira [26]3 years ago

8 0

Answer: My friend told me it is 79

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125 I will mark branliest if right<br> x 35<br> ₋₋₋₋₋₋₋₋₋₋₋₋₋<br> + ₋₋₋₋₋₋₋₋₋₋₋₋₋<br> ₋₋₋₋₋₋₋₋₋₋₋₋₋

Mariana [72]

Answer:

125

× 35

_________

625

375

_______

4 375

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_______

4400

Step-by-step explanation:

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3 years ago

If C is the current value and S is the starting value, write an

LekaFEV [45]

Is there a diagram with this question?

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3 years ago

What is the equation of the line that passes through (-2,-3) and is perpenicular to 2X -3Y= 6

Rufina [12.5K]

You should first of all rewrite the equation in slope intercept form. You should get y=2/3x-2. The perpendicular slope is the opposite reciprocal, in this case -3/2. You will use this slope and the point to write your equation in point slope form and then clean this up to get it into slope intercept form. You should get y=-3/2x-6

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3 years ago

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guapka [62]

Answer:

Did you got my answer

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3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago