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3 years ago

Which of the following is true of a radius of a circle?

Mathematics

2 answers:

Sav [38]3 years ago

6 0

The second one is correct (Radius is the distance from the center to any point on the circle)

nikdorinn [45]3 years ago

6 0

Answer:

2.)

Step-by-step explanation:

The radius of any circle is the distance from the EXACT center, to any point of the circle.

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Leni [432]

I'm just going to take a chance and say 1.A, and 2.D.

8 0

3 years ago

Find the inverse function g(x) of the function f(x)=2x+1

choli [55]

Answer:

The inverse function is f(x) = (x - 1)/2

Step-by-step explanation:

To find the inverse of any function, start by switching the x and f(x) values.

f(x) = 2x + 1

x = 2f(x) + 1

Now solve for the new f(x). The result will be your inverse function.

x = 2f(x) + 1

x - 1 = 2f(x)

(x - 1)/2 = f(x)

6 0

3 years ago

Identify all sets to which the number belongs.

ra1l [238]

Option B

-0.249851765 is a irrational number

<em><u>Solution:</u></em>

Given number is -0.249851765

We have to classify the number

Let us first understand about irrational, real, whole, integer and rational numbers

<h3><u>Integers:</u></h3>

An integer is a whole number (not a fractional number) that can be positive, negative, or zero. Examples of integers are: -5, 1, 5, 8, 97, and 3,043

Every integer can be expressed as a decimal, but most numbers that can be expressed as a decimal are not integers. If all the digits after the decimal point are zeroes, the number is an integer. If there are any non-zero digits after the decimal point, the number is not an integer.

Thus -0.249851765 has non zero digits after decimal point. So it is not a integer

<h3><u>Whole numbers:</u></h3>

Whole numbers are positive numbers, including zero, without any decimal or fractional parts. Negative numbers are not considered "whole numbers."

But the given number -0.249851765 is negative number. So it is not a whole number

<h3><u>Natural numbers:</u></h3>

A natural number is an integer greater than 0. Natural numbers begin at 1 and increment to infinity: 1, 2, 3, 4, 5, etc. Natural numbers are also called "counting numbers" because they are used for counting.

A decimal is a natural number if it is non-negative and the only digits after its decimal points are zero

So the given number -0.249851765 is a negative number and so it is not a natural number

<h3><u>Rational numbers:</u></h3>

A rational number is a number that can be expressed as a fraction (ratio) in the form $\frac{p}{q}$ where p and q are integers and q is not zero.

The rational numbers includes all positive numbers, negative numbers and zero that can be written as a ratio (fraction) of one number over another.

When a rational number fraction is divided to form a decimal value, it becomes a terminating or repeating decimal.

So -0.249851765 is not a rational number

<h3><u>Irrational number:</u></h3>

An irrational number is real number that cannot be expressed as a ratio of two integers.

When an irrational number is expressed in decimal form, it goes on forever without repeating

So the given number -0.249851765 is irrational number

We can conclude that:

-0.249851765 is a irrational number, So Option B is correct

3 0

3 years ago

Simplify:<br><br> {(-8)^-4 ÷ 2^-8}^2

Aleksandr-060686 [28]

Answer:

ok so first we have to do whats in the brackets then we have to do to exponits so first

(-0.00024414062divided by 0.00390625)^2

-0.06249999872^2

-0.00390624984

Hope This Helps!!!

8 0

2 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

4 years ago