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Nimfa-mama [501]
3 years ago
6

Find three consecutive integers such that the sum of the first and twice the second is equal to the third plus four.

Mathematics
1 answer:
salantis [7]3 years ago
8 0
Our three consecutive integers are

First: x
Second: x + 1
Third: x + 2

Here is the equation you need:

x + 2(x + 1) = (x + 2) + 4

Take it from here.
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What is the range of the function in the graph?
Verdich [7]

The range of the function in the graph given can be expressed as: D. 20 ≤ s ≤ 100.

<h3>What is the Range and Domain of a Function?</h3>

All the possible set of x-values that are plotted on the horizontal axis (x-axis) are the domain of a function. In order words, they are the inputs of the function.

On the other hand, all the corresponding set of x-values that are plotted on the vertical axis (y-axis) are the range of a function. In order words, they are the outputs of the function.

In the graph given below that shows a function, s is plotted on the vertical axis (y-axis), and its values starts from 20, up to 100. The range can be said to be all values of s that are from 20 to 100.

Therefore, the range of the function in the graph given can be expressed as: D. 20 ≤ s ≤ 100.

Learn more about functions on:

brainly.com/question/15602982

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3 0
2 years ago
If the measure of the angle m is twice the measure of its complement if the given angle is 40 degrees, What is the measure of an
sladkih [1.3K]
C. Because complementary angles are two angles whose sum equals 90 degrees so do 90-40=50 then 50x2= 100
4 0
3 years ago
Help I need this done ASAP anyone know how to solve this
ludmilkaskok [199]
Equation~ 5b + (2b - 4) + (3b -6)= 180.
4 0
3 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
Descriptive Measures Instructions: Use the following data set to find the sample statistics for the following data set. (Round a
iren2701 [21]

Answer:

(1) N or n = 20

(2)  = 43.4

(3) σ = 9.78 and s = 10.04.

Step-by-step explanation:

8 0
3 years ago
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