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3 years ago

PLEASE HELPPPPP.

1 answer:

3 0

1. You'd divide 98,000,000 by 750,000 and get about 131 trees per acre ("population" density).

2. 131 trees/acre x 750,000 = 98,250,000trees.
Then 292,000,000 divided by98,250,000, which is 2.9 Round up to 3 boxes/tree.

Hope I helped. Hope this makes sense.

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What is 12/25 divided by 4

UNO [17]

In order to find the answer to this, you must first turn 4 into a fraction (4/1)

12/25 divided by 4/1 is the same as 12/25 multiplied by 1/4, because when dividing fractions, you must use the reciprocal of the divisor and multiply.

12/25 x 1/4 = 12/100 = 6/50 = 3/25

Your answer should be 3/25<span />

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3 years ago

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The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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2 years ago

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The table displays the amount of time, in minutes, Jordan rode his bike for five days. How many total hours did Jordan ride his

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Answer:98.9276

Step-by-step explanation:

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What is the mean of the set of numbers?

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The mean of your question would be -2.5

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How do I Solve this problem?

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3 years ago

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