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3 years ago

1. A chess set has 32 pieces, 16 black and 16 white. In each color, there

Mathematics

1 answer:

USPshnik [31]3 years ago

6 0

Answer:

I believe the answer is A.

Step-by-step explanation:

I don't really know how to really explain it tho

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Kesara santiago purchased a riding mower for $2,746.00. she received a $49.95 rebate from the manufacturer and a $38.95 rebate f

andrew11 [14]

A rebate means that the buyer receives back a portion of the original price of the item. So in this case, the rebate prices must be deducted from the original price to get the final price, that is:

final price = $2,746 - $49.95 - $38.95

<span>final price = $2,657.1</span>

8 0

4 years ago

Super easy plz answer all for 20 points just adding

sergejj [24]

7/8+9/10=71/40=1 31/40
1/4+2/3=11/12
1/2+1/4 is different from 1/2+1/3 because
1/2+1/4= 2/4+1/4= 5/1 =1 1/4
1/2+1/3=3/6+2/6=5/6

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3 years ago

How much will I get if I get a 8 percent of $108000

iren [92.7K]

Answer:

8,640

Step-by-step explanation:

EASY PERIOD

5 0

4 years ago

I need help asap

In-s [12.5K]

You know the formula for calculating the lateral area of a cylinder with a radius r and height h is S=2*π*r*h
So we have S= 2*π*6*20= 240π
Have fun

6 0

4 years ago

Read 2 more answers

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