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Gelneren [198K]
3 years ago
6

a psychologist contends that the number of facts of a certain type that are remembered after t hours is given by the following f

unction. ​f(t)equals startfraction 85 t over 99 t minus 85 endfraction find the rate of change at which the number of facts remembered is changing after 1 hour and after 10 hours.
Mathematics
1 answer:
laiz [17]3 years ago
4 0

Answer:

At t=1, Rate of Change=-36.86

At t=10 hours, Rate of Change =-0.0088

Step-by-step explanation:

The function which describes the number of facts of a certain type which are remembered after t hours is given as:

f(t)=\frac{85t}{99t-85}

To determine the Rate of Change at the given time, we first look for the derivative of f(t).

Applying quotient rule:

f^{'}(t)=\frac{-7225}{{\left( 85 - 99\,t\right) }^{2}}

At t=1

f^{'}(1)=\frac{-7225}{(85-99)^{2}}

=-36.86

At t=10 hours

f^{'}(10)=\frac{-7225}{(85-99(10))^{2}}

=-0.0088

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One-third of a number t is equal to 7.<br> Write the word sentence as an equation then solve
Artemon [7]

Answer:

21

Step-by-step explanation:

One third of the number can be represented by 1/3t.

To make an equation, set this equal to 7.

1/3t = 7

Then, solve for t by dividing each side by 1/3:

1/3t = 7

t = 21

So, the number is 21.

6 0
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У + 2х = 7<br> у = 3х – 3<br> What is the answer to this
Neko [114]
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3 years ago
If 30,000 cm2 of material is available to make a box with a square base and an open top, what is the largest possible volume (in
Cloud [144]

Answer:

The largest possible volume of the box is 2000000 cubic meters.

Step-by-step explanation:

The volume (V), in cubic centimeters, and surface area (A_{s}), in square centimeters, of the box with a square base are described below:

A_{s} = l^{2}+h\cdot l (1)

V = l^{2}\cdot h (2)

Where:

l - Side length of the base, in centimeters.

h - Height of the box, in centimeters.

By (2), we clear h within the formula:

h = \frac{V}{l^{2}}

And we apply in (1) and simplify the resulting expression:

A_{s} = l^{2}+ \frac{V}{l}

A_{s}\cdot l = l^{3}+V

V = A_{s}\cdot l -l^{3} (3)

Then, we find the first and second derivatives of this expression:

V' = A_{s}-3\cdot l^{2} (4)

V'' = -6\cdot l (5)

If V' = 0 and A_{s} = 30000\,cm^{2}, then we find the critical value of the side length of the base is:

30000-3\cdot l^{2} = 0

3\cdot l^{2} = 30000

l = 100\,cm

Then, we evaluate this result in the expression of the second derivative:

V'' = -600

By Second Derivative Test, we conclude that critical value leads to an absolute maximum. The maximum possible volume of the box is:

V = 30000\cdot l - l^{3}

V = 2000000\,cm^{3}

The largest possible volume of the box is 2000000 cubic meters.

4 0
2 years ago
14/2 three equivalent ratios
Anna71 [15]

Answer:

Three equivalent ratios of  \frac{14}{2} are \frac{7}{1}, \frac{21}{3} and \frac{28}{8}.

Step-by-step explanation:

We are given a fraction \frac{14}{2}.

We need to find the three equivalent ratios/fractions of \frac{14}{2}.

The common factor of 14 and 2 is 2.

So, dividing top and bottom by 2, we get

14÷2 = 7 and 2÷2 that is \frac{7}{1}.

Multiplying \frac{7}{1} fraction by a common number 3 in top and bottom, we get

7 × 3 = 21

1 × 3 = 6

So, we get another equivalent fraction \frac{21}{6}.

Multiplying \frac{7}{1} fraction by a common number 3 in top and bottom, we get

7 × 4 = 28

1 × 4 = 4

So, we get another equivalent fraction \frac{28}{4}.

Therefore, three equivalent ratios of  \frac{14}{2} are \frac{7}{1}, \frac{21}{3} and \frac{28}{4}.

6 0
3 years ago
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