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QveST [7]
2 years ago
15

A sign next to a roller coaster at an amusement park says, “You must be at least 60 inches tall to ride.” Noah is happy to know

that he is tall enough to ride.
Noah’s friend is 2 inches shorter than Noah. Can you tell if Noah’s friend is tall enough to go on the ride? Explain or show your reasoning.

Mathematics
2 answers:
Anna11 [10]2 years ago
8 0

Answer:

it depends, if Noah is taller than 60 by more than 2 inches his friend can ride but if he is exactly 60 inches then his friend would not be able to ride

Step-by-step explanation:

atroni [7]2 years ago
7 0

Answer:

if Noah is 60 inches tall (5ft) and noah's friend is 2 inches shorter than he is 58inches tall, so no he won't be able to. but Noah could be 61 inches, but his friend is still to short.

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Nutka1998 [239]
Point slope form is:

y - y_{1} = m(x - x_{1})

With y_{1} and x_{1} being the coordinate points, and m being the slope.

So the coordinate points are (4, -5), and the slope is -6 in this case.

So just substitute everything in, to get:

y - ( -5) = -6(x - 4)

Or

y + 5 = -6 (x - 4)
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Rewrite 2x2x2 using an exponent.
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2 years ago
Assuming that the equation defines x and y implicitly as differentiable functions xequals​f(t), yequals​g(t), find the slope of
Doss [256]

Answer:

\dfrac{dx}{dt} = -8,\dfrac{dy}{dt} = 1/8\\

Hence, the slope , \dfrac{dy}{dx} = \dfrac{-1}{64}

Step-by-step explanation:

We need to find the slope, i.e. \dfrac{dy}{dx}.

and all the functions are in terms of t.

So this looks like a job for the 'chain rule', we can write:

\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx} -Eq(A)

Given the functions

x = f(t)\\y = g(t)\\

and

x^3 +4t^2 = 37 -Eq(B)\\2y^3 - 2t^2 = 110 - Eq(C)

we can differentiate them both w.r.t to t

first we'll derivate Eq(B) to find dx/dt

x^3 +4t^2 = 37\\3x^2\frac{dx}{dt} + 8t = 0\\\dfrac{dx}{dt} = \dfrac{-8t}{3x^2}\\

we can also rearrange Eq(B) to find x in terms of t , x = (37 - 4t^2)^{1/3}. This is done so that \frac{dx}{dt} is only in terms of t.

\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

\dfrac{dx}{dt} = \dfrac{-8t}{3(37 - 4t^2)^{2/3}}\\\dfrac{dx}{dt} = \dfrac{-8(3)}{3(37 - 4(3)^2)^{2/3}}\\\dfrac{dx}{dt} = -8

now let's differentiate Eq(C) to find dy/dt

2y^3 - 2t^2 = 110\\6y^2\frac{dy}{dt} -4t = 0\\\dfrac{dy}{dt} = \dfrac{4t}{6y^2}

rearrange Eq(C), to find y in terms of t, that is y = \left(\dfrac{110 + 2t^2}{2}\right)^{1/3}. This is done so that we can replace y in \frac{dy}{dt} to make only in terms of t

\dfrac{dy}{dt} = \dfrac{4t}{6y^2}\\\dfrac{dy}{dt}=\dfrac{4t}{6\left(\dfrac{110 + 2t^2}{2}\right)^{2/3}}\\

we can find the value of this derivative using t = 3, and plug that value in Eq(A).

\dfrac{dy}{dt} = \dfrac{4(3)}{6\left(\dfrac{110 + 2(3)^2}{2}\right)^{2/3}}\\\dfrac{dy}{dt} = \dfrac{1}{8}

Finally we can plug all of our values in Eq(A)

but remember when plugging in the values that \frac{dy}{dt} is being multiplied with \frac{dt}{dx} and NOT \frac{dx}{dt}, so we have to use the reciprocal!

\dfrac{dy}{dx} = \dfrac{dy}{dt} .\dfrac{dt}{dx}\\\dfrac{dy}{dx} = \dfrac{1}{8}.\dfrac{1}{-8} \\\dfrac{dy}{dx} = \dfrac{-1}{64}

our slope is equal to \dfrac{-1}{64}

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3 years ago
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arlik [135]

Answer:

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