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Maru [420]
3 years ago
11

Determine the x- and y-intercepts of the graph of y=−12x−4 . Then plot the intercepts to graph the equation.

Mathematics
2 answers:
a_sh-v [17]3 years ago
7 0

Answer:

<em><u>X-Intercept</u></em>: \frac{4}{-12}

<em><u>Y-Intercept</u></em>: -4

Step-by-step explanation:

Hello, great question. These types are questions are the beginning steps for learning more advanced Algebraic Equations.

The X and Y intercepts show at what points the function values cross the X and Y axis. We can solve for the X-intercept by substituting the value of y for 0 and then solving for x. Like so...

0=-12x-4 .... add 4 on both sides

4=-12x ... divide both sides by -12

\frac{4}{-12} = -x

Now we can see that the X-Intercept is \frac{4}{-12} . Now we substitute 0 for x and solve for the Y-intercept.

y=-12(0)-4

y= -4

Finally we can see that the Y-intercept is -4

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Morgarella [4.7K]3 years ago
6 0
X-intercept: -0.3
y-intercept: -4
slope -12
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2 years ago
Light travels about 180 million kilometers in 10 minutes. How far does it travel in one minute? How far does it travel in one se
astraxan [27]

Answer:

18 million kilometers in one min. and 0.3 million kilometers in one second

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
A cube has an edge length of 9 meters. What is its volume, in cubic meters?
rjkz [21]
<h3>Answer:  729 cubic meters</h3>

Work Shown:

volume = side*side*side

volume = 9*9*9

volume = 729

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2 years ago
(-72)/(-6)<br> Awnser <br> Plz&amp;Thx
KIM [24]
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8 0
3 years ago
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Find the measurements (the length L and the width W) of an inscribed rectangle under the line with the 1st quadrant of the x &am
Leni [432]

The question is incomplete. Here is the complete question.

Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -\frac{3}{4}x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.

Answer: L = 1; W = 9/4; A = 2.25;

Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:

A = x.y

A = x(-\frac{3}{4}.x + 3)

A = -\frac{3}{4}.x^{2}  + 3x

To maximize, we have to differentiate the equation:

\frac{dA}{dx} = \frac{d}{dx}(-\frac{3}{4}.x^{2}  + 3x)

\frac{dA}{dx} = -3x + 3

The critical point is:

\frac{dA}{dx} = 0

-3x + 3 = 0

x = 1

Substituing:

y = -\frac{3}{4}x + 3

y = -\frac{3}{4}.1 + 3

y = 9/4

So, the measurements are x = L = 1 and y = W = 9/4

The maximum area is:

A = 1 . 9/4

A = 9/4

A = 2.25

6 0
3 years ago
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