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mestny [16]
3 years ago
10

Write an equation in slope-intercept form of the line that passes through the given point and is parallel to the graph of the gi

ven equation.
(2-2) ;y=-x-2
A.y=-2x
B.y=2x
C.y=1/2x
D.y=-x
I have no idea how to work this out, I have a lot of other questions that are like this so just an answer isn't really going to help me out so much. Thanks for all help given!
Mathematics
1 answer:
Marrrta [24]3 years ago
5 0
 y = x - 2.....the slope here is 1. A parallel line will have the same slope

y = mx + b
slope(m) = 1
(2,-2)...x = 2 and y = -2
now we sub and find b, the y int
-2 = 1(2) + b
-2 = 2 + b
-2 - 2 = b
-4 = b

so ur parallel line is : y = x - 4

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Help
MissTica
I think it might be C
7 0
3 years ago
Read 2 more answers
Solve for x.
S_A_V [24]

Answer:

x = {-2, 2/3}

Step-by-step explanation:

2x = -4

x = -2

3x = 2

x = 2/3

x = {-2, 2/3}

3 0
3 years ago
Solve for x: 0.75x-0.25=3.5
exis [7]

Answer:

x=5

Step-by-step explanation:

3.5=0.75x-0.25

add 0.25 to 3.5

3.75=0.75x

divide 3.75 by 0.75

x=5

4 0
3 years ago
The daily cost of producing x high performance wheels for racing is given by the following​ function, where no more than 100 whe
algol [13]

Answer:

The production would be 25 wheels,

Lowest average cost is $ 123.75

Step-by-step explanation:

Given cost function,

C(x) = 0.09x^3 - 4.5x^2 + 180x

Where,

x = number of wheel,

So, the average cost per wheel,

A(x) = \frac{C(x)}{x}=\frac{0.09x^3-4.5x^2 + 180x}{x}=0.09x^2 - 4.5x + 180

Differentiating with respect to x,

A'(x) = 0.18x - 4.5

Again differentiating with respect to x,

A''(x) = 0.18

For maxima or minima,

A'(x) = 0

0.18x - 4.5 = 0

0.18x = 4.5

\implies x = \frac{4.5}{0.18}=25

For x = 25, A''(x) = positive,

i.e. A(x) is maximum at x = 25.

Hence, the production would be 25 wheels for the lowest average cost per​ wheel.

And, lowest average cost,

A(x) = 0.09(25)² - 4.5(25) + 180 = $ 123.75

5 0
3 years ago
FURTHER MATHEMATICS Use determinants to solve the systems of equation:
maw [93]

Answer:

Step-by-step explanation:

\left\{\begin{array}{ccc}2x+1y+2z&=&13\\1x+1y-2z&=&8\\1x+2y+1z&=&11\\\end{array}\right.\\\\\\\Delta=\left| \begin{array}{ccc}2&1&2\\1&1&-2\\1&2&1\end{array}\right| =2*\left| \begin{array}{ccc}1&\frac{1}{2}&1\\1&1&-2\\1&2&1\end{array}\right| =2*\left| \begin{array}{ccc}1&\frac{1}{2}&1\\0&\frac{1}{2}&-3\\0&1&3\end{array}\right| =2*(\frac{3}{2}+3)=9\\\\

\Delta_1=\left| \begin{array}{ccc}13&1&2\\8&1&-2\\11&2&1\end{array}\right| =2*\left| \begin{array}{ccc}13&1&2\\8&1&-2\\\frac{11}{2}& 1&\frac{1}{2}\end{array}\right| =2*\left| \begin{array}{ccc}13&1&2\\-5&0&-4\\\frac{-5}{2}& 0&\frac{5}{2}\end{array}\right| \\\\=2*(-1)*(\frac{-25}{2}-\frac{20}{2}) =45\\

\Delta_2=\left| \begin{array}{ccc}2&13&2\\1&8&-2\\1&11&1\end{array}\right| \\\\\\=\left| \begin{array}{ccc}3&21&0\\3&30&0\\1&11&1\end{array}\right| \\\\\\=1*(90-63) =27\\

\Delta_3=\left| \begin{array}{ccc}2&1&13\\1&1&8\\1&2&11\end{array}\right| \\\\\\=\left| \begin{array}{ccc}0&-1&-3\\0&-1&-3\\1&2&11\end{array}\right| \\\\\\=0\\

\left\{\begin{array}{ccc}x=\dfrac{\Delta_1}{\Delta}=\dfrac{45}{9}=5\\\\y=\dfrac{\Delta_2}{\Delta}=\dfrac{27}{9}=3\\\\z=\dfrac{\Delta_3}{\Delta}=\dfrac{0}{9}=0\\\end{array}\right.

8 0
3 years ago
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