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3 years ago

HELP ME PLEASEEEEEEEEEEEEEEEEEE

Mathematics

1 answer:

Mice21 [21]3 years ago

6 0

Answer:

y = 3/2x + 15

Step-by-step explanation:

change f(x) to 'y='

interchange 'x' and 'y' then solve for 'y':

y = 2/3x - 10

x = 2/3y - 10

x+10 = 2/3y

multiply each side by 3/2 to get:

y = 3/2x + 15

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Step-by-step explanation:

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3 years ago

the length of a rectangle is 4 cm less than 5 times its width its perimeter is 160 cm find its length and width

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Let x= width
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3 years ago

Assume m<2 = 35°. Find the measure of all of the other

soldi70 [24.7K]

m<1 = 145 | Supplementary
m<3 = 35 | Vertical
m<4 = 145 | Supplementary
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3 years ago

Read 2 more answers

A model rocket is launched with an initial velocity of 200 ft per second. The height h, in feet, of the rocket t seconds after t

Solnce55 [7]

Answer:

2.10 s, 10.40 s.

Step-by-step explanation:

We know that the height of the rocket is given by the function:

$h=-16t^2+200t$

We are asked to find the time for which the height of the rocket will be 350 ft. So, for that moment, we know the height but we don't know the time; however, we know that the equation can help us to find the time, doing h=350:

$350=-16t^2+200t$

The last is a quadratic equation, which can be put in the form $at^2+bt+c=0$ and solved applying the formula:

$t=\frac{-b+-\sqrt{b^2-4ac} }{2a}$

So, let's put the equation on the form $at^2+bt+c=0$ adding $16t^2$ and subtracting $200t$ to each side of the equation; the result is:

$16t^2-200t+350=0$

So, we note that a=16, b=-200, and c=350.

Then,

$t_1=\frac{200-\sqrt{200^2-4*16*350} }{2*16}=2.10$

$t_2=\frac{200+\sqrt{200^2-4*16*350} }{2*16}=10.40$

According to the equation, that are the times for which the height will be 350 ft; that is because the rocket is going to ascend and then to fail again to the ground.

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3 years ago

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