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Tems11 [23]
3 years ago
10

What are the intercepts of the line 3x – 15y = 60?   A. x-intercept: –4; y-intercept: 20   B. x-intercept: –3; y-intercept: 15  

C. x-intercept: 20; y-intercept: –4   D. x-intercept: 5; y-intercept: 20
Mathematics
1 answer:
AnnZ [28]3 years ago
8 0
X intercept is when y = 0
3x -15(0) = 60
3x = 60
x = 20

y intercept is when x=0
3(0) -15y = 60
-15y = 60
y = -4

Answer is C
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The similarities between constructing a perpendicular line through a point on a line and constructing a perpendicular through a point off a line include:

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<h3>What are perpendicular lines?</h3>

Perpendicular lines are defined as two lines that meet or intersect each other at right angles.

In this case, both methods involve making a 90-degree angle between two lines and the methods determine a point equidistant from two equidistant points on the line.

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Find the solution to the system of equations. <br> x+y+z=23 <br> y+z=14 <br> z=9
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<span>A) x+y+z=23
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Use the slope formula to find the slope of the line in the graph shown above.
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An object is propelled upward from the top of a 300 foot building. The path that the object takes as it falls to the ground can
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Answer:

As per the statement:

The path that the object takes as it falls to the ground can be modeled by:

h =-16t^2 + 80t + 300

where

h is the height of the objects and

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At t = 0 , h = 300 ft

When the objects hit the ground, h = 0

then;

-16t^2+80t+300=0

For a quadratic equation: ax^2+bx+c=0         ......[1]

the solution for the equation is given by:

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On comparing the given equation with [1] we have;  

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then;

t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}

t= \frac{-80\pm \sqrt{6400+19200}}{-32}

t= \frac{-80\pm \sqrt{25600}}{-32}  

Simplify:

t = -\frac{5}{2} = -2.5 sec and t = \frac{15}{2} = 7.5 sec

Time can't be in negative;

therefore, the time it took the object to hit the ground is 7.5 sec

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