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3 years ago

A large square carpet is initially x inches long and x inches wide.

Mathematics

1 answer:

ddd [48]3 years ago

5 0

(x - 10) × 2
You have to multiply it by 2 so that you get the area

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If he starts at 1 and jumps 3 units to the right, then where is he on the number line? How far away from zero is he?

Ilya [14]

Answer:

If you mean he starts at 1 on the Y axis (vertical line), then the answer is (1, 3). If you mean he starts at 1 on the X axis (horizontal line), then the answer is (0, 4)

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A watch company is developing packaging for its new watch. The designer uses octagons with a base area of 30 in2 and rectangles

bezimeni [28]

Answer:

It's (C) 240in3

Step-by-step explanation:

Volume of the prototype is 250 in.³

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3 years ago

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Pleease help me with this math work

OleMash [197]

The first one because the left bound is included [ and the right one is not )

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3 years ago

Can someone help me complete this worksheet please and thank you it’s all multiple choice

7nadin3 [17]

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3 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