Mathematics, 09.06.2020 23:57 EMQPWE

Going from point A to point B, the cheetah traveled at an average rate of 70 mph. Returning to point A, the cheetah traveled at an average rate of 40 mph. Can we say that this cheetah’s average rate was 55 mph? The average rate for the trip is equal to the total distance traveled divided by the total time traveled. The following equations represent the distance traveled on each leg of the trip. First leg of trip: d=r_1 t_1 Second leg of trip: d=r_2 t_2 Write an equation for the average rate for the trip. Remember, the cheetah runs from point A to point B and back to point A.

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Answer from: tishfaco5000

The average rate of a body is the total distance travelled divided by the total time.

The cheetah runs at an average rate of 50.91mph, not 55mph

Let

$d \to$ distance

$r \to$ average rate

$t \to$ time

Given that:

d = r_1t_1 = r_2t_2

So, we have:

r_1 = 70

r_2 = 40

The average rate (r) is calculated as follows:

$r = \frac{Total\ Distance (D)}{Total\ Time (T)}$

Where:

D=d + d

D = r_1t_1 + r_2t_2

and

T =t_1 + t_2

So, the average rate is:

$r =\frac{r_1t_1 + r_2t_2}{t_1 + t_2}$

Recall that:

D = r_1t_1 + r_2t_2

r_1t_1 = r_2t_2

Make t_2 the subject

$t_2 = \frac{r_1t_1}{r_2}$

Substitute values for r_1 and r_2

$t_2 = \frac{70t_1}{40}$

So, we have:

$r =\frac{r_1t_1 + r_2t_2}{t_1 + t_2}$

$r = \frac{70t_1 + 40 \times \frac{70t_1}{40}}{t_1 + \frac{70t_1}{40}}$

$r = \frac{70t_1 + 70t_1}{t_1 + \frac{70t_1}{40}}$

$r = \frac{140t_1}{t_1 + \frac{70t_1}{40}}$

Factor out t1

$r= \frac{140t_1}{t_1(1 + \frac{70}{40})}$

$r = \frac{140}{(1 + \frac{70}{40})}$

$r = \frac{140}{\frac{40+70}{40}}$

$r= \frac{140}{\frac{110}{40}}$

Rewrite as:

$r = \frac{140 \times 40}{110}$

r = 50.91mph

Hence, the cheetah's average rate is 50.91mph, not 55mph

Read more about distance, average rates and time at:

link

Answer from: billyeyelash

No.

$\text{Average rate} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{r_{1}t_{1} + r_{2}t_{2}}{t_{1} + t_{2}}\\\\= \text{50.91 mi/h}$

Step-by-step explanation:

First leg: d = r₁t₁

Second leg: d = r₂t₂

r₁t₁ = r₂t₂

Total distance: 2d = r₁t₁ + r₂t₂

Total time: t = t₁ + t₂

1. Equation for average rate

$\text{Average rate} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{r_{1}t_{1} + r_{2}t_{2}}{t_{1} + t_{2}}$

2. Average rate

Since r₁t₁ = r₂t₂,

$t_{1} = \dfrac{r_{2}t_{2}}{r_{1}} = \frac{40}{70}t_{2} = \frac{4}{7}t_{2}\\\\\text{Average rate} = \dfrac{2r_{2}t_{2}}{\frac{4}{7}t_{2} + t_{2}} = \dfrac{2 \times 40t_{2}}{\frac{4t_{2}+ 7t_{2}}{7}}= 80t_{2} \times\frac{7}{11t_{2}} = \dfrac{560}{11}\\\\= \textbf{50.91 mi/h}$

We cannot say truthfully that the average rate is 55 mi/h.

Answer from: Quest

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function $y=f(x)$ the average rate of change from point $a(a,f(a))$ to point $b(b,f(b))$ can be calculated as

$\text{average rate of change}=\dfrac{f(b)-f(a)}{b-a}.$

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,

$\text{average rate of change}=\dfrac{2-8}{10-2}=\dfrac{-6}{8}=-\dfrac{3}{4}.$