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2 years ago

Laura will go for a run during her lunch break if the temperature is between 60∘F

Mathematics

1 answer:

balu736 [363]2 years ago

8 0

Answer:

D(t) = 3 + 0.0(80 - t)

Step-by-step explanation:

The average of speed of Laura in miles per hour is given by:

S(t) = 6 + 0.1(80 - t) Equation 1

where, t is the temperature in degrees Fahrenheit.

The distance D, Laura covers at x miles per hour is given as:

D(x) = 0.5x Equation 2

We need to find the expression that models the distance that Laura runs in terms of the temperature "t"

The "x" in Equation 1 represents the average speed of Laura in miles per hour. S(t) in Equation 1 also represent the speed of Laura in miles per hour. So, we can replace x by S(t) in Equation 2 and generate an equation of Distance in terms of temperature "t" as shown below:

D(S(t)) = 0.5 (6 + 0.1(80-t))

D(t) = 3 + 0.0(80 - t)

This expression models the distance that Laura runs in 30 minutes given that it is t∘F outside at the start of her run.

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Zack borrowed $1,087 for 12 months at 11% interest. if he must pay 11.00 per $100, what is the total amount of money he will rep

Naddika [18.5K]

Answer:

The correct option is d.

Step-by-step explanation:

It is given that Zack borrowed $1,087 for 12 months at 11% interest. It means he must pay 11.00 per $100.

Formula of simple interest:

$I=\frac{P\times r\times t}{100}$

Where P is principal amount, r is rate of interest in percentage and t is time in years.

$I=\frac{1087\times 11\times 1}{100}$

$I=119.57$

The interest paid by Zack is $1,087.

The total amount of money he will repay is

$1,087+119.57=1,206.57$

Therefore option d is correct.

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2 years ago

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Adriano runs a website that helps writers write better stories. Every month, Adriano receives a subscription fee of \$5$5dollar

jenyasd209 [6]

Answer:

$1002516150

Step-by-step explanation:

200200200 + 404040 - 101010 = 200503230

200503230 x 5 = 1002516150

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1 year ago

Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

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2 years ago

Which of the following notations correctly describe the end behavior of the polynomial graphed below

Irina18 [472]

The answer is B
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2 years ago

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Clara lends half her collection of formal dresses, d, to her sister, Susan. Clara then buys four more dresses. If she has 12 now

Verdich [7]

Answer:

Step-by-step explanation:

a:2+4 =12

a:2=12-4

a:2=8

a=8 ×2

a=16 dresses

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