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3 years ago

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A suit regularly cost 375.95 cents Bob buys it on sale for 268.35 cents how much does he save

1 answer:

Stolb23 [73]3 years ago

7 0

Hello, all we need to do in this case to subtract the amount of money Bob paid for the suit from the amount of money the suit regularly costs. So:

375,95 - 268,35 = 107,6

Bob saves 107,6$.

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A cylinder has a radius of 12 m and a height of 9 m. What is the exact volume of the cylinder? 108π m³ 216π m³ 972π m³ 1296π m³

Mumz [18]

V= πr^2h
V = π * 12^2 * 9
= 1296π

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Simplify the expression, Please help!!

Serhud [2]

For the first one it would be C.

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(My teacher asked who our favorite Artist was and used each artist to each person who liked that artist. Awesome right? But- don

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3520. just multiply all 3 lwh

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SVEN [57.7K]

Answer:

Step-by-step explanation:

1l ........8.5 miles

x l .......6 miles

-----------------------

x=6*1/8.5

x=0.70 l

2*0.7=1.4 l petrol/day ( to work and come back home)

5*1.4=7 l/week ( 5 days works in a week)

7*1.26=8.82 L /week

8.82>8.5

The petrol costs more

So the answer is NO

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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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