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zhenek [66]
3 years ago
6

A beach ball rolls off a cliff and onto the beach. The height, in feet, of the beach ball can be modeled by the function h(t)=64

–16t^2, where t represents time, in seconds.
What is the average rate of change in the height, in feet per second, during the first 1.25 seconds that the beach ball is in the air?
Mathematics
2 answers:
True [87]3 years ago
6 0

You are given h(t) = 64–16t^2.

You need to find h(1.25).

Set up:

h(1.25) = 64 - 16(1.25)^2

Take it from here.

AVprozaik [17]3 years ago
6 0

Answer:

Step-by-step explanation:

Given the height of a beach ball rolling off a cliff as a function of time

h = 64 — 16t²

We want to find the average change in height when t = 1.25s

The rate of change of height is the differentiation of the height I.e

∆h/∆t → dh/dt

Therefore

h = 64 — 16t².

Therefore

Differential of a constant is zero

dh/dt = —32t

At t = 1.25s

dh/dt = —32(1.25)

dh/dt = —40 ft/s

So, the height is reducing at a rate of 40ft in every seconds.

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12. MP Persevere with Problems Write the function rule for each function.
densk [106]

Answer: 4. 78,020,008 4. Seventy-eight million, twenty thousand, eight

5. 20,011,012 5. 20,000,000 + 11,000 + 12

6. 50,000 7. 900 8. 70,000,000 9. 30,000,000

Step-by-step explanation:

7 0
2 years ago
A can in the shape of a right circular cylinder is required to have a volume of 500 cubic centimeters. The top and bottom are ma
borishaifa [10]

Answer:

The answer to the question is

The total cost C of the material as a function of the radius r of the cylinder is

0.6912·r² + 800/r Dollars.

Step-by-step explanation:

To solve the question, we note that

The area of the top and bottom combined  = 2·π·r²

The area of the sides = 2·π·r·h

and the volume = πr²h = 500 cm²

Therefore height = 500/(πr²)

Substituting the value of h into the area of the side we have

Area of the side = 2πr·500/(πr²) = 1000/r

Therefore total area of can = Area of top + Area of bottom + Area of side

Whereby the cost of the can = 0.11×Area of top +0.11×Area of bottom +0.8×Area of side

Which is equal to

0.11×2×π×r²+ 0.8×1000/r = 0.6912·r² + 800/r

The cost of the can is $(0.6912·r² + 800/r)

3 0
3 years ago
Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.
DENIUS [597]

Answer:

The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

The probability that a randomly chosen plate contains the sub-string HI is 0.002548  approximately.

Step-by-step explanation:

Consider the provided information.

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.

The numbers of ways in which three letters followed by three digits can be chosen is: 26\times 26\times 26 \times 10 \times 10 \times10

26^3\times 10^3=17576000

The numbers of ways in which two letters followed by four digits can be chosen is: 26\times 26\times 10 \times 10 \times 10 \times10

26^2\times 10^4=6760000

Hence, the total number of ways are 17576000 + 6760000 = 24336000

Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.

Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are: 26^2=676

The probability that a randomly chosen plate contains the number 2222 is:

\frac{676}{24336000} \approx 0.000028

Part (B)

The number of ways in which chosen plate contains the sub-string HI:

If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:

26\times 1\times10^3+1\times 26\times10^3

If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:

1\times 10^4

Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:

1\times 10^4+26\times 1\times10^3+1\times 26\times10^3

62000

From part (A) we know that the total number of ways to chose a number plate is 24336000.

The probability that a randomly chosen plate contains the sub-string HI is:

\frac{62000}{24336000} \approx 0.002548

7 0
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Olenka [21]

Answer:

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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