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horrorfan [7]
3 years ago
8

Write the equation of the line that is parallel to the graph of y=-4x -9, and whose y-intercept is 3.

Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

y = -4x +3

Step-by-step explanation:

the only thing you need from the equation is the slope, which is -4

y = -4x +3

You might be interested in
How to you solve -15=y-62
sdas [7]
First move the (Y) to the other side of the equal sign,
- y - 15 = - 62
Then subtract add (15) on both sides
- y = - 47
Since a variable can't be negative just divide by (-1) on both sides,
\frac{ - y}{ - 1} = \frac{ - 47}{ - 1}
The answer would be,
y = 47
Hope this helped
:D
7 0
3 years ago
I need help with math
Keith_Richards [23]

\frac{1}{5}  +  \frac{3}{4}  \times  \frac{3}{5}  =  \\

\frac{1}{5}  +  \frac{9}{20}  =  \\

\frac{4}{20}  +  \frac{9}{20}  =  \\

\frac{13}{20}  \\

4 0
3 years ago
A countertop is in the shape of a trapezoid. The lengths of the bases are 70.5 and 65.5 inches long. The height of the counterto
den301095 [7]
The answer is 1224. You do 70.5 + 65.5, divided by two, then you take that answer and multiply by 18 to get 1224
8 0
3 years ago
A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

  • the angle of elevation is Ф = 45° and
  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0
2 years ago
Can anyone help me???
lbvjy [14]

Answer:

I think it's J.

Step-by-step explanation:

I don't know how to explain it. •-• Sorry!

7 0
3 years ago
Read 2 more answers
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