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4 years ago

8

In a plane, if one of two _____ lines is perpendicular to another line, the other of the two lines is perpendicular to that line

. A. perpendicular B. parallel C. congruent D. equal

1 answer:

MA_775_DIABLO [31]4 years ago

7 0

Answer:

A

Step-by-step explanation:

If a parallel line is perpendicular to the other line, the line parallel to the first line is also perpendicular to that line.

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Y+4−1=18 i need help plz

d1i1m1o1n [39]

Answer:

y=15

Step-by-step explanation:

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3 years ago

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Take 39/50 simplify into a fraction

Nat2105 [25]

Answer:

This fraction is in the simplest form already.

Step-by-step explanation:

This fraction is in the simplest form already. Here is why.

Result in decimals: 0.78

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3 years ago

A particular style of sunglasses costs the retailer $70 $⁢70 per pair. At what price should the retailer mark them so he can sel

Alexxx [7]

After the 10% discount on marked price (P), the selling price is ...

... P×(100% -10%) = 0.90P

After a markup off 35% on cost, the value is

... $70×(100% +35%) = 1.35×$70 = $94.50

The retailer wants these two values to be equal:

... 0.90P = $94.50

... P = $94.50/0.90 = $105.00 . . . . . divide by the coefficient of P

The marked price should be $105.00.

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3 years ago

A triangle can have one right angle.<br> a. True<br> b. False

iragen [17]

True because a triangle must have 3 angles that add up to 180 degrees and, since a right angle is 90 degrees, it cannot have 2 right angles

3 0

4 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