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3 years ago

13

Film travels through a projector at the rate of 170 feet per minute. How many feet of flim are in a motion picture that last 120

minutes

1 answer:

Vaselesa [24]3 years ago

6 0

120x170=20400

hope it helps

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F(x)=x^(2)+3x+4 is this quadratic, linear, or neither

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2 years ago

4. Gloria the grasshopper is working on her hops.

aivan3 [116]

The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola that describes the path of her jump is $\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

The given parameters are:

$\mathbf{Height = 20}$

$\mathbf{Length = 28}$

<em>Assume she starts from the origin (0,0)</em>

The midpoint would be:

$\mathbf{Mid = \frac 12 \times Length}$

$\mathbf{Mid = \frac 12 \times 28}$

$\mathbf{Mid = 14}$

So, the vertex of the parabola is:

$\mathbf{Vertex = (Mid,Height)}$

Express properly as:

$\mathbf{(h,k) = (14,20)}$

A point on the graph would be:

$\mathbf{(x,y) = (28,0)}$

The equation of a parabola is calculated using:

$\mathbf{y = a(x - h)^2 + k}$

Substitute $\mathbf{(h,k) = (14,20)}$ in $\mathbf{y = a(x - h)^2 + k}$

$\mathbf{y = a(x - 14)^2 + 20}$

Substitute $\mathbf{(x,y) = (28,0)}$ in $\mathbf{y = a(x - 14)^2 + 20}$

$\mathbf{0 = a(28 - 14)^2 + 20}$

$\mathbf{0 = a(14)^2 + 20}$

Collect like terms

$\mathbf{a(14)^2 =- 20}$

Solve for a

$\mathbf{a =- \frac{20}{14^2}}$

$\mathbf{a =- \frac{20}{196}}$

Simplify

$\mathbf{a =- \frac{5}{49}}$

Substitute $\mathbf{a =- \frac{5}{49}}$ in $\mathbf{y = a(x - 14)^2 + 20}$

$\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

Hence, the equation of the parabola that describes the path of her jump is $\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}$

See attachment for the graph

Read more about equations of parabola at:

brainly.com/question/4074088

7 0

2 years ago

PLS HELP!!!!!!!!!!!!!

stepladder [879]

First, multiply the top equation by 2 and bottom by -3.
60x+60y=900
-60x-75y=-1035
————————
-15y=-135
—————-
-15=-15
y=9

So the answer should be 9.

6 0

3 years ago