An item is regularly priced at

If a cup of coffee has temperature 95∘C in a room where the temperature is 20∘C, then, according to Newton's Law of Cooling, the

muminat

Answer:

61°C

Step-by-step explanation:

Newton's Law of cooling gives the temperature -time relationship has:

T (t) = 20 + 75 е⁻(t/50)-------------------------------------------------------- (1)

where Time is in minutes (min) & Temperature in degree Celsius (°C)

During the first half hour, t = 30 mins

Substituting into (1)

T = 20 + 75 е⁻(30/50)

= 20 + 75(0.5488)

= 20 + 41.16

= 61.16°C

≈ 61°C

4 0

3 years ago

If a population has a mean of 128 and samples of size 50 from this population have a standard deviation in their sample means of

Angelina_Jolie [31]

Answer is B, but please study for your future you don’t need help from others trust me just study for 2 hours a day that will help you to develop some skills and make brainiest that’s mean you will develop some brain cells that you

7 0

3 years ago

Use the Remainder Theorem to find the remainder when P(x) = x^4 – 9x^3 – 5x^2 – 3x + 4 is divided by x + 3.

kirza4 [7]

P(x) = (x + 3)Q(x) + R

<span> P(-3) = (0)Q(x) + R </span>

<span>i P(-3) = R </span>

<span>Now, P(-3) = (-3)⁴ - 9(-3)³ - 5(-3)² - 3(-3) + 4
</span>So

<span>P(-3) = 81 + 243 - 45 + 9 + 4 </span>
<span>=. 292
hope it helps</span>

6 0

3 years ago

Read 2 more answers

Ilona is at a friend's home that is several miles from her home. She starts walking at a constant rate in a straight line toward

larisa [96]

Complete question is;

Ilona is at a friends home that is several miles from her home. She starts walking at a constant rate in a straight line toward her home.

The expression -2t + 10 gives the distance, in miles, that Ilona is from her home after t hours. Select True or False for each of the following.

1. The friends home is 10 miles from Ilonas home

2. The distance Ilona has walked away from her friends home after t hours is given by the absolute value of -2t

3.The negative sign in the term indicates that Ilona is walking toward her home

4. Ilona is walking at a rate of 10 miles per hour

Answer:

Statements 1 & 2 are correct.

Step-by-step explanation:

We are told that the expression -2t + 10 gives the distance, in miles, that Ilona is from her home after t hours.

Now, we know that distance = speed x time

Now, from the expression -2t + 10

It's clear that 2t is a distance value.

This means that the speed used in going home is 2 miles/hr

This means that at t hours, she has walked 2t miles from her friends home.

Thus, statement 2 is true because an absolute value of -2t will be 2t.

Since 2t is subtracted from 10 in the expression and 2t is the distance she has walked from her friends house at Time, t. It means that 10 miles is the distance from her friends house to her house.

Thus, statement 1 is also true

7 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago