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Otrada [13]
2 years ago
8

The temperature of a weather station was 12.8 °F at midnight and fell 1.4 °F each hour for six hours. Which of the following sol

utions are viable for (h,T) where h is the number of hours after midnight and T is the temperature in degrees Fahrenheit?
Mathematics
2 answers:
Len [333]2 years ago
3 0

Answer:

Let's try to find a linear relation like:

T(h) = a*h + b

where a is the slope and b is the y-intercept.

h is the number of hours after midnight.

T is the temperature at the time defined by h.

We know that at midnight, the temperature is 12.8°F.

At midnight, we have h = 0, then:

T(0) = a*0 + b = 12.8°F

       b = 12.8°F

Now we know that our function is:

T(h) = a*h + 12.8°F

We also know that the temperature fell 1.4 °F each hour for six hours.

Then the slope will be -1.4°F

We can write the linear relationship as:

T(h) = -1.4°F*h + 12.8°F     (for 0 ≤ h ≤ 6)

Where we have a restriction in the possible values of h, because we know that this model only works for six hours after midnight,

Alja [10]2 years ago
3 0

Answer:

Step-by-step explanation:

The answer is (5, 5.8)

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d1i1m1o1n [39]

Answer:

About the x axis

V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}

About the y axis

V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}

About the line y=8

V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}

About the line x=2

V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi

Step-by-step explanation:

For this case we have the following functions:

y = 2x^2 , y=0, X=2

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on  y from 0 to 8.

We can find the area like this:

A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4

And we can find the volume with this formula:

V = \int_{a}^b A(x) dx

V= 4\pi \int_{0}^2 x^4 dx

V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}

About the y axis

For this case we need to find the function in terms of x like this:

x^2 = \frac{y}{2}

x = \pm \sqrt{\frac{y}{2}} but on this case we are just interested on the + part x=\sqrt{\frac{y}{2}} as we can see on the second figure attached.

We can find the area like this:

A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})

And we can find the volume with this formula:

V = \int_{a}^b A(y) dy

V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy

V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)

And we can find the volume with this formula:

V = \int_{a}^b A(x) dx

V= \pi \int_{0}^2 64-32x^2 +4x^4 dx

V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}

And we can find the volume with this formula:

V = \int_{a}^b A(y) dy

V= \frac{\pi}{2} \int_{0}^8 y dy

V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi

6 0
3 years ago
Heelp meeee !!!!!!!!!!
Katen [24]
The function is in fact a polynomial and the degree of the polynomial is 7.
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I've got a math test wish me luck​
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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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2 years ago
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