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liraira [26]
3 years ago
15

The graph below has the same shape. what is the equation of the graph?

Mathematics
1 answer:
Nataly_w [17]3 years ago
3 0

C; A and B are incorrect because the shift is not vertical, it is horizontal. This is because 3 is in parenthesis or a 'grouping' (same thing if it is in an exponent) and is being modified. Since the graph is being translated down 3 units, D is incorrect since it is a translation up 3 units. The answer is C.

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2 years ago
At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce
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Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then \frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }.  By the Mean Value Theorem, there is a number c such that 0 < c with v'(c)=60 \:{\frac{mi}{h^2}}. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 60 \:{\frac{mi}{h^2}}.

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

  1. f is continuous on the closed interval [a, b].
  2. f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

f'(c)=\frac{f(b)-f(a)}{b-a}

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then v(0 \:h) = 30 \:{\frac{mi}{h} } and v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} } (note that 20 minutes is 20/60=1/3 of an hour), so the average rate of change of v on the interval [0 \:h, \frac{1}{3} \:h] is

\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in (0 \:h, \frac{1}{3} \:h) at which v'(c)=60 \:{\frac{mi}{h^2}}.

c is a time time between 2:00 and 2:20 at which the acceleration is 60 \:{\frac{mi}{h^2}}.

4 0
3 years ago
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tekilochka [14]

Answer:

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Step-by-step explanation:

The closest you can get to 42 with that equation and it has to use consecutive intergers is 41 and these numbers get you there.

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