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Helga [31]
3 years ago
9

Amir drove from Jerusalem down to the lowest place on Earth, the Dead Sea, descending at a rate of 12 meters per minute. He was

at sea level after 30 minutes of driving. Graph the relationship between Amir's altitude relative to sea level (in meters) and time (in minutes).

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
6 0
<h2>Answer:</h2>

First of all let's write the slope-intercept form of the equation of a line, which is:

y=mx+b \\ \\ Where: \\ \\ m: \ slope \\ \\ b: \ y-intercept

So we just need to find m \ and \ b to solve this problem.

Moreover, this problem tells us that Amir drove from Jerusalem down to the lowest place on Earth, the Dead Sea, descending at a rate of 12 meters per minute. So this rate is the slope of the line, that is:

m=-12

Negative slope because Amir is descending. So:

y=-12x+b

To find b, we need to use the information that tells us that he was at sea level after 30 minutes of driving, so this can be written as the point (30,0). Therefore, substituting this point into our equation:

y=-12x+b \\ \\ 0=-12(30)+b \\ \\ 0=-360+b \\ \\ \therefore b=360

Finally, the equation of Amir's altitude relative to sea level (in meters) and time (in minutes) is:

\boxed{y=-12x+360}

Whose graph is shown bellow.

pantera1 [17]3 years ago
4 0

Answer:

-12+360

Step-by-step explanation:

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Step-by-step explanation:

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\text{Given that,}\\\\f(x) =  \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0}  \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\

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<h3>2)</h3>

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