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skad [1K]
2 years ago
13

Simplify: 5a + 1r + 2a + 4r

Mathematics
1 answer:
Marina CMI [18]2 years ago
4 0

Answer:

7a+5r

Step-by-step explanation:

5a + 1r + 2a + 4r

add like terms

7a+5r

hope that helps

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What is the vertical distance from the origin to the point (2,9)
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9

Step-by-step explanation:

9.

y coordinate gives the vertical distance

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3 years ago
Find the missing values<br> 3:4=6:__=9:_=12:_=15:_
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Step-by-step explanation:

6:8=9:12=12:16=15:20

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Step-by-step explanation:

3rd one

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3 years ago
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7 0
2 years ago
State the equation of the line<br> that is perpendicular to 4x-3y=10<br> through the point (-2,4)
Lynna [10]

The equation of the line  that is perpendicular to 4x - 3y = 10  through the point (-2,4) is y-4=\frac{-3}{4}(x+2)

<u>Solution:</u>

Given, line equation is 4x – 3y = 10  

We have to find a line that is perpendicular to 4x – 3y = 10 and passing through (-2, 4)

Now, let us find the slope of the given line,  

\text { Slope of a line }=\frac{-\mathrm{x} \text { coefficient }}{\mathrm{y} \text { coefficient }}=\frac{-4}{-3}=\frac{4}{3}

We know that, slope of a line \times slope of perpendicular line = -1

\begin{array}{l}{\text { Then, } \frac{4}{3} \times \text { slope of perpendicular line }=-1} \\\\ {\rightarrow \text { slope of perpendicular line }=-1 \times \frac{3}{4}=-\frac{3}{4}}\end{array}

Now, slope of our required line = \frac{-3}{4} and it passes through (-2, 4)

<em><u>The point slope form is given as:</u></em>

\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on the line. }} \\\\ {\text { Here in our problem, } m=-\frac{3}{4}, \text { and }\left(x_{1}, y_{1}\right)=(-2,4)} \\\\ {\text { Then, line equation } \rightarrow y-4=-\frac{3}{4}(x-(-2))}\end{array}

y-4=\frac{-3}{4}(x+2)

Hence the equation of line is found out

6 0
2 years ago
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