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frosja888 [35]
3 years ago
10

Solve for x - Pre algebra

Mathematics
2 answers:
aniked [119]3 years ago
7 0
In this case you would set it up as
6x + 4x + 130 = 180
(180 represents the degree of the triangle)

Add up like terms
10x + 130 = 180
Subtract 130 to the other side (180-130)

10x = 50
Divide 10 to the other side ( 50/10)
.
x = 5
Hope this was clear :)
Montano1993 [528]3 years ago
7 0

The sum of the measures of the interior angles of a triangle is 180 degrees.

Make the three angles add up to 180 in an equation.

6x + 130 + 4x = 180

Combine like terms.

10x + 130 = 180

Subtract 130 from both sides.

10x = 50

Divide both sides by 10.

x = 5 is your answer.

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Answer:

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Step-by-step explanation:

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The length of a rectangle is 3 less than twice the width. Write a function rule for the perimeter in terms of w, the width
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What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S
Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is y=\frac{1}{3} x+2

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3  

And as the line passing through (-3,1) and is  perpendicular to 3x + y = -8, product of slopes of two line will be -1  as lies are perpendicular.

Let required slope = x  

\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}

So we need to find the equation of a line whose slope is \frac{1}{3} and passing through (-3,1)

Equation of line passing through (x_1 , y_1) and having lope of m is given by

\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)

\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}

Substituting the values we get,

\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}

Hence the required equation of line is found using slope intercept form

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