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andrew11 [14]
3 years ago
15

Use the following graph to estimate the rate of change of the function at x=0.6 using the points (0,0) and (1,−0.5)

Mathematics
1 answer:
svetoff [14.1K]3 years ago
4 0

Answer:

The rate of change of the function at x = 0.6 is approximately -0.72

Step-by-step explanation:

The given information from the graph are;

Points (0, 0) and (1, -0.5)

We have that the graph is that of a cubic function, therefore;

f(x) = ax³ + bx² + cx + d

At f'(x) at  (0, 0) and (1, -0.5) = 0 gives;

3ax² + 2bx + c = 0

∴ c = 0

Also

ax³ + bx² + cx + d = 0

d = 0

We have

3a(1)² + 2b(1) + 0 = 0

3·a + 2·b = 0..............(1)

a(1)³ + b(1)² + c(1) + d = -0.5

a + b + 0×(1) + 0 = -0.5

a + b = -0.5..........(2)

Multiplying equation (2) by 2 and subtracting it from equation (1) gives;

2 × (a + b = -0.5) = 2·a + 2·b = -1

3·a + 2·b - (2·a + 2·b) = 0 - (-1) = 0 + 1 = 1

a = 1

From

a + b = -0.5, we have;

1 + b = -0.5 = 0

b = -0.5 - 1 = -1.5

The equation becomes

f(x) = x³ - 1.5·x²

The rate of change of the function at x = 0.6 is therefore given as follows;

f'(x) = 3 × x² - 2 × (1.5)·x = 3·x² - 3·x

At x = 0.6, we have;

f'(0.6) = 3·(0.6)² - 3·(0.6)

f'(0.6) = 3×0.6² - 3×0.6 = -0.72

The rate of change of the function at x = 0.6 ≈ -0.72

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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
At what position on the number line is the red dot located?
Lostsunrise [7]

Answer:

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Step-by-step explanation:

2 times 2 is 4

3 times 3 is 9

7 is the only option between 4 and 9

(this was more the easy method lol but hope this helps)

5 0
3 years ago
10 times The sum of half a numberand six is eight
GuDViN [60]

The numerical form of this worded expression is:

10(1/2n + 6)=8

4 0
3 years ago
Your closet has 5 shirts for every 4 sweaters. Your closet has 15 shirts. How many sweaters are in your closet?
Mnenie [13.5K]
Since it is a 5:4 it would be 15:12
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2 years ago
Read 2 more answers
Upon descent, an airplane is 20,000 feet above the ground. The air traffic control tower is 200 feet tall. It is determined that
PIT_PIT [208]

Answer:

Base length of plane from control tower = 73,880 feet (Approx.)

Step-by-step explanation:

Given:

Height of plane from ground = 20,000 feet

Height of control tower = 200 feet

Angle of elevation from control tower = 15°

Find:

Base length of plane from control tower

Computation:

Height of plane from control tower = 20,000 - 200

Height of plane from control tower = 19,800 feet

Tan θ = Perpendicular / Base

Tan15 = 19,800 / Base length of plane from control tower

0.268 = 19,800 / Base length of plane from control tower

Base length of plane from control tower = 73,880 feet (Approx.)

6 0
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