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Kryger [21]
3 years ago
12

The changing speed of a car is modeled by the function S(t) = –4t + 35, where t is time in seconds. Interpret the model.

Mathematics
1 answer:
dangina [55]3 years ago
6 0

Answer:

The car has an initial speed of 35 units and is slowing down by 4 units per second.

Step-by-step explanation:

The changing speed of a car is modeled by the function :

s(t)=-4t+35

t is time in seconds

Initial speed means at t = 0

s(0) = 35 units

Here the change in speed is negative, it means that the car is slowing down by this rate.

So, the correct answer is : The car has an initial speed of 35 units and is slowing down by 4 units per second.

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Answer:

1. (28-2)x

2.(12+6)x

3.(15+1)x

Step-by-step explanation:

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3 years ago
Which of the following expressions is the conjugate of a complex number with −5 as the real part and 4i as the imaginary part?
makvit [3.9K]
"−5 − 4i" is the one among the following expressions given in the question that <span>is the conjugate of a complex number with −5 as the real part and 4i as the imaginary part. The correct option among all the options that are given in the question is the third option. I hope that this is the answer that has helped you.</span>
3 0
3 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
2 years ago
Someone arranges three 24-karat gold squares (ofequalthickness)as shown below, and offers you a choice. You can
Katarina [22]

Answer:

idk

Step-by-step explanation:

4 0
3 years ago
43 points, 86 for brainliest! Answer correct!! Be wise! According to the image attached, whats the answer?
Snezhnost [94]

Answer:

p = 32\:  \: in

a = 49 {in}^{2}

Step-by-step explanation:

perimeter \\  = 8 + 8+8+8 \\  = 32 \:  \: in

area \\   1)3 \times 8 = 24 \\ 2)5 \times 5 = 25 \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 49 {in}^{2}

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0
3 years ago
Read 2 more answers
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