Find the value of the expression 8a +b, if a=3 and b= 1 a) 23 b) 29 c) 30 d) 25

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

6 0

2 years ago

Read 2 more answers

Solve each proportion

max2010maxim [7]

13. x= 27/4 or 6 and 3/4
14. x= 27
15. x= 191/7 or 27 and 2/7
16. x= 83/15 or 5 8/15
the key to solving these is cross-multiplying and finding a common denominator, you can use photomath if you want to see the solving steps, hope this helps!

3 0

3 years ago

Nick went canoeing on the river he traveled 1 mile in 17 minutes.at this rate,how far will he travel in an hour and 25 minutes

dem82 [27]

9514 1404 393

Answer:

5 miles

Step-by-step explanation:

An hour and 25 minutes is 85 minutes. Using the given information we can find the distance d with the proportion ...

$\dfrac{\text{distance}}{\text{time}}=\dfrac{d}{85\text{ min}}=\dfrac{1\text{ mi}}{17\text{ min}}\\\\d=\dfrac{85\text{ min}}{17\text{ min}}(1\text{ mi)}\qquad\text{multiply by $85$ min}\\\\d=5(1\text{ mi})\qquad\text{simplify}\\\\d=\boxed{5\text{ mi}}$

In 1 hour and 25 minutes, Nick will travel 5 miles.

6 0

3 years ago

What is the product (x-3)(2x^2-5x + 1)

Reika [66]

F '(x) = 2(3x2 + 5) + 6x(2x - 1) ... + 3), s(x) = x - 4. Use the product rule table. below to get: r'(x) = 20x + 11. s'(x) = 1 ... f'(x) = (x - 4)(20x + 11) - (5x - 2)(2x + 3)/(x - 4)2.First, we will distribute 2x to (x + 5), then we will distribute 3. ... 5x. That is,. multiplying binomials. 6x − x= 5x. Example 2. Multiply (3x − 1)(x + 2). Answer. 3x2 + 5x

4 0

3 years ago

F(x) = -3x-3 g(x) =2x^3+2 Find f(6) and g(-3)

Ket [755]

usa pothomath (no me acuerdo como se escribe)

4 0

3 years ago