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1 year ago

10

The cheerleaders held a car wash to raise money for new uniforms they spent $15 on soap and eight dollars on sponges rags in a b

ucket it rain the day of the car wash so the cheerleaders only made $27

2 answers:

garik1379 [7]1 year ago

4 0

Answer: The cheerleaders made 4 dollars of profit

Step-by-step explanation:

There isn't a question but I suppose you are trying to solve how much the cheerleaders made. 27-15-8 = 4

chubhunter [2.5K]1 year ago

3 0

Answer:

The answer will be $6

Step-by-step explanation:

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Hi I need this question please asap.

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Ike Jehoshaphat isn’t oso los

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3 years ago

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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Indicate the equation of the given line in standard form. The line through the midpoint of and perpendicular to the segment join

lianna [129]

The midpoint is (3,-1)

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3 years ago

A fan is marked up 40% on the original price the original price was 20$ what is the new price of the fan before sales tax

Art [367]

Answer:

28 dollars is the new price.

Step-by-step explanation:

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3 years ago

What is the value of x and y?

kramer

Angles in a triangle add up to 180 degrees.

All angles in an equilateral triangle have equal values

Both angle x and y equal 60 degrees.

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2 years ago