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3 years ago

In the diagram, which pair of angles are vertical angles?

Mathematics

1 answer:

Alexxx [7]3 years ago

6 0

Answer:

<6&<8

Step-by-step explanation:

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Solve the inequality. Show your work. |r + 3| ≥ 7

poizon [28]

R could equal any number, but it has to be higher than 4, because 3+4=7.

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3 years ago

An avid traveler is investigating whether travel websites differ in their pricing. She chooses a random sample of 32 hotel rooms

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The answer is D beachside trust

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2 years ago

State whether the following statements are true or false, and explain your reasoning.

Masja [62]

Answer:

Step-by-step explanation:

a) A square is a rectangle. True

Reason: Property of Rectangle: (i) Opposite sides are equal and parallel. (ii) Each angle is 90 (iii) Diagonals are equal and bisect each other.

A square hold all these properties. A square is a rectangle.

b) A polygon with 21 sides has 432 possible diagonals. False

Reason: Number of diagonals of a polygon = $\frac{n(n-3)}{2}$

n --> number of sides of a polygon.

= $\frac{21*18}{2}$ = 189 diagonals

c) All three angles in an Isosceles triangle are equal. False

Reason: In an isosceles triangle, two angles are equal.

If three angles are equal, then that is an equilateral triangle.

d) The measures of the exterior angles of a nonagon, a nine-sided figure, have a sum of 360°. True.

Reason: The sum of measures of the exterior angles of any polygon is 360

6 0

3 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

Help please!! <br> What is the domain of the function shown in the graph?

diamong [38]

Answer:

x ≥ - 1

Step-by-step explanation:

4 0

3 years ago