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2 years ago

8

Jerald jumped from a bungee tower. If the equation that models his height, in feet, is h = –16t2 + 729, where t is the time in s

econds, for which interval of time is he within 104 feet above the ground?

1 answer:

Veseljchak [2.6K]2 years ago

5 0

I believe it isssssss 6.5

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2 years ago

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Kate's swim coach informs her that she only has 35% of her workout left. Kate has swum 26 laps so far.

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2 years ago

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Licemer1 [7]

The slope of all parallel lines are equal.

The slope of a line in the form $y=mx+c$ is m.

1. The given equation is: $y=4x+2$ . The slope of this line is $m=4$ . The slope of the line parallel to this line is also 4.

2. The given equation is $y=\frac{2}{7}x+1$ . The slope of this line is $m=\frac{2}{7}$ .

The slope of the line parallel to it is also $\frac{2}{7}$ .

3. The next equation is $y+9x=13$ . The slope intercept form is $y=-9x+13$ .

The slope of this is $m=-13$ . The line parallel to it also has slope which is -13.

4. The given equation is $y=-\frac{1}{2}x+4$ . The slope of this line and the line parallel to it is $m=-\frac{1}{2}$

5. The equation is 6x+2y=4. We simplify to get: 3x+y=2.

The slope-intercept form is $y=-3x+2$ . The line parallel to this line has slope $m=-3$

6. Assuming the line is $y=-3x+9$ , then the slope is $m=-3$

7. Assuming the line is $-5x+5y=4$ , then the slope-intercept form is $y=x+\frac{4}{5}$ and the slope of the line parallel to this line is m=1

8. The equation is 9x-5y=4. The slope intercept form is $y=\frac{5}{9}x-\frac{4}{9}$ . The slope of the line parallel to this line is $m=\frac{5}{9}$

9. The given equation is $-x+3y=6$ . The slope intercept form is $y=\frac{1}{3}x+2$ . The line parallel to this line also has slope $m=\frac{1}{3}$ .

10. Assuming the given line is 6x-7y=10, then $y=\frac{6}{7}x-\frac{10}{7}$ and the line parallel to it has slope $m=\frac{6}{7}$

11. Assuming the line is $y-x=-3$ , then $y=x-3$ and the slope of the line parallel to it is $m=1$

12. The given line is 3x-5y=6. This implies that $y=\frac{3}{5}x-\frac{6}{5}$

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2 years ago

Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

6 0

3 years ago

Find the product.<br> (3b + 6at)(b - at)

adoni [48]

Answer:

3bat + 3b² - 6a²t²

Step-by-step explanation:

First you have to expand it to get;

3b(b - at) + 6at(b - at)

Then you can now multiply.

3b² - 3bat + 6bat - 6at²

Group like terms

6bat - 3bat + 3b² -6at²

3bat + 3b² - 6a²t²

4 0

3 years ago

Read 2 more answers