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2 years ago

Write the equation of the line perpendicular to y = 5/3x + 1 that passes through the point (- 8, 2)

Mathematics

1 answer:

xenn [34]2 years ago

4 0

Answer:

y = - $\frac{3}{5}$ x - $\frac{14}{5}$

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = $\frac{5}{3}$ x + 1 ← is in slope- intercept form

with slope m = $\frac{5}{3}$

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{\frac{5}{3} }$ = - $\frac{3}{5}$ , thus

y = - $\frac{3}{5}$ x + c ← is the partial equation

To find c substitute (- 8, 2) into the partial equation

2 = $\frac{24}{5}$ + c ⇒ c = 2 - $\frac{24}{5}$ = - $\frac{14}{5}$

y = - $\frac{3}{5}$ x - $\frac{14}{5}$ ← equation of perpendicular line

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X-2y=18 2x+y=6 parallel perpendicular or neither

sergey [27]

<h3>Therefore they are perpendicular.</h3>

Step-by-step explanation:

A equation of line is

y =mx +c

Here the slope of the line is m.

Given equations are

x - 2y = 18

⇔-2y = -x +18

$\Leftrightarrow y =\frac{1}{2} x -9$ ............(1)

and 2x + y = 6

⇔y = -2x +6 ............(2)

Therefore the slope of equation (1) is $(m_1)$ = $\frac{1}{2}$

Therefore the slope of equation (2) is $(m_2)$ = -2

If two lines are perpendicular, when we multiply their slope we get -1.

therefore,

$m_1. m_2$ = $\frac{1}{2}$ . (-2) = -1

Therefore they are perpendicular.

4 0

2 years ago

Hey all!

Rainbow [258]

Answer:

65°

Step-by-step explanation:

<u>Use the law of cosines:</u>

cos C = (25² + 27² - 28²)/(2*25*27)
cos C = 0.4222
m∠C = arccos 0.4222
m∠C = 65°

8 0

2 years ago

HUUURRRYYYY PLEASEEEEE!!!!!!!What is the approximate volume of a cone with a height of 6 mm and radius of 18 mm? Use 3.14 to app

bulgar [2K]

V=pir^2(h/3)
V= 3.14x18^2x6/3
V= about <span>2034.72</span>

3 0

3 years ago

Explain how to multiply the following whole numbers 21 x 14

Lesechka [4]

Answer:

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Step-by-step explanation:

Given

$21\:\times \:14$

Line up the numbers

$\begin{matrix}\space\space&2&1\\ \times \:&1&4\end{matrix}$

Multiply the top number by the bottom number one digit at a time starting with the ones digit left(from right to left right)

Multiply the top number by the bolded digit of the bottom number

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

Multiply the bold numbers: 1×4=4

$\frac{\begin{matrix}\space\space&2&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&\space\space&4\end{matrix}}$

Multiply the bold numbers: 2×4=8

$\frac{\begin{matrix}\space\space&\textbf{2}&1\\ \times \:&1&\textbf{4}\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the top number by the bolded digit of the bottom number

$\frac{\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&8&4\end{matrix}}$

Multiply the bold numbers: 1×1=1

$\frac{\begin{matrix}\space\space&\space\space&2&\textbf{1}\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&\space\space&1&\space\space\end{matrix}}$

Multiply the bold numbers: 2×1=2

$\frac{\begin{matrix}\space\space&\space\space&\textbf{2}&1\\ \space\space&\times \:&\textbf{1}&4\end{matrix}}{\begin{matrix}\space\space&\space\space&8&4\\ \space\space&2&1&\space\space\end{matrix}}$

Add the rows to get the answer. For simplicity, fill in trailing zeros.

$\frac{\begin{matrix}\space\space&\space\space&2&1\\ \space\space&\times \:&1&4\end{matrix}}{\begin{matrix}\space\space&0&8&4\\ \space\space&2&1&0\end{matrix}}$

adding portion

$\begin{matrix}\space\space&0&8&4\\ +&2&1&0\end{matrix}$

Add the digits of the right-most column: 4+0=4

$\frac{\begin{matrix}\space\space&0&8&\textbf{4}\\ +&2&1&\textbf{0}\end{matrix}}{\begin{matrix}\space\space&\space\space&\space\space&\textbf{4}\end{matrix}}$

Add the digits of the right-most column: 8+1=9

$\frac{\begin{matrix}\space\space&0&\textbf{8}&4\\ +&2&\textbf{1}&0\end{matrix}}{\begin{matrix}\space\space&\space\space&\textbf{9}&4\end{matrix}}$

Add the digits of the right-most column: 0+2=2

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

Therefore,

$\begin{matrix}\space\space&\textbf{2}&\textbf{1}\\ \times \:&1&\textbf{4}\end{matrix}$

________

$\frac{\begin{matrix}\space\space&\textbf{0}&8&4\\ +&\textbf{2}&1&0\end{matrix}}{\begin{matrix}\space\space&\textbf{2}&9&4\end{matrix}}$

6 0

3 years ago

Write the set of points from −6−6 to 33 but excluding −2−2 and 33 as a union of intervals:

Softa [21]

In this question, we have to write the set of points from -6 to 3 but excluding -2 and 3 as a union of intervals .

For union, we use U.

For the points that we have to exclude , we put parenthesis on there side that is ().

Therefore the required interval form is

$[-6,-2)U(-2,3)$

As we see that () are used with -2 and 3 . And that's the required interval form .

5 0

3 years ago