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GaryK [48]
2 years ago
7

Find the area of the polygon

Mathematics
1 answer:
AlekseyPX2 years ago
7 0

Answer:

The answer is 383m^{2} .

To find the area of a rectangle, we multiply the length of the rectangle by the width of the rectangle.

Please Mark Brainliest If This Helped!

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Pleeeeaaaaase help me
Elenna [48]

Step-by-step explanation:

measure all the sides then multiply you will get the answer

4 0
2 years ago
Newton's law of cooling is:
Mnenie [13.5K]

Answer:

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

\frac{du}{dt}= -k (u-T)

We can reorder the expression like this:

\frac{du}{u-T} = -k dt

We can use the substitution w = u-T and dw =du so then we have:

\frac{dw}{w} =-k dt

IF we integrate both sides we got:

ln |w| = -kt +C

If we apply exponential in both sides we got:

w = e^{-kt} *e^c

And if we replace w = u-T we got:

u(t)= T + C_1 e^{-kt}

We can also express the solution in the following terms:

u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}

For this case we know that k =-0.15 hr since w ehave a cooloing, T_{i}= 70 F, T_{amb}=11F, we have this model:

u(t) = (70-11) e^{-0.15t} +11

And if we want that the temperature would be 32F we can solve for t like this:

32 = 59 e^{-0.15 t} +11

21=59 e^{-0.15 t}

\frac{21}{59} = e^{-0.15 t}

If we apply natural logs on both sides we got:

ln (\frac{21}{59}) =-0.15 t

t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr

So it would takes approximately 6.9 hours to reach 32 F.

7 0
3 years ago
WILL GIVE BRAINLIEST PLEASE HELP
kirill [66]

Answer:

25%=7  75%=21  100%=28  125%= 35  150%=42

7 0
2 years ago
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Write the equation of a vertical line that goes through (2,-1)
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The equation would be ... Y = 2x + (-5)

6 0
3 years ago
please explain to me I don't understand why x -2 less than/ equal to 0 but x + 4<0...I don't understand why the inequalities
krok68 [10]
If you look at the image you can see that it says “x - 2 is greater than/equal to 0
3 0
3 years ago
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