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3 years ago

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Alex returns a bike to a friends house from his own house in 1/2 hour. He walks back over the same route in 1.5 hours. If the av

erage speed on the bike is 8 mph faster than his walking speed, how far does alex live from his friend

1 answer:

ira [324]3 years ago

4 0

Answer:

Step-by-step explanation:

Let x represent his average walking speed.

If the average speed on the bike is 8 mph faster than his walking speed,it means that his biking speed is (x + 8) mph

Distance = speed × time

Alex returns a bike to a friends house from his own house in 1/2 hour. This means that the distance covered is

0.5(x + 8)

He walks back over the same route in 1.5 hours. This means that the distance covered is

1.5x

Since the distance covered is the same, then

1.5x = 0.5(x + 8)

1.5x = 0.5x + 4

1.5x - 0.5x = 4

x = 4 mph

Therefore, the distanceof Alex's house from his friend's house is

4 × 1.5 = 6 miles

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Tiana can clear a football field of debris in 3 hours. Jacob can clear the same field in 2 hours. If they decide to clear the fi

Lapatulllka [165]

Let us determine each of their rates of work:

Since Tiana can clear the field in 3 hours, meaning she can do 1/3 of the work per hour.

In the same way, Jacob can do 1/2 of the work per hour.

Now, what would be the rate of work if they were working together? Let's look at it like this:

Hours to complete the job:
Tiana = 3
Jacob = 2
Together = t

Work done per hour:
Tiana = 1/3
Jacob = 1/2
Together = 1/t

If you add their labor together:

1/3 + 1/2 = 1/t
5/6 = 1/t
t = 6/2 = 1.2

Together, they can clear the field in 1.2 hours.

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4 years ago

Read 2 more answers

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

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4 years ago

[1x-6]>1 what are the solutions

a_sh-v [17]

Answer:

x > 7

(I think this is Right but not sure yet )

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3 years ago

19. There were 20 drinks in a cooler.

ella [17]

For this case we have that the complete question is:

There were 20 drinks in a cooler. Joey drank 15% of the drinks. Sandra drank 1/5 of the drinks. Hannah drank 1/4 of the drinks. Tammy drank 30% of the drinks. How many drinks are left in the cooler? a 0 b 2 c 5 d 9

We propose a rule of three:

20 ---------> 100%

x ------------> 15%

Where the variable "x" represents the amount of drinks equivalent to 15%.

$x = \frac {15 * 20} {100}\\x = 3$

So, Joey drank 3 drinks.

On the other hand, we have that Sandra drank $\frac {1} {5}$ of the drinks:

$20 * \frac {1} {5} = \frac {20} {5} = 4$

Thus, Sandra drank 4 drinks.

In addition, Hannah drank $\frac {1} {4}$ of the drinks:

$20 * \frac {1} {4} = 5$

So Hannah drank 5 drinks.

Finally, Tammy drank 30% of the drinks:

20 ---------> 100%

y ------------> 30%

Where the variable "y" represents the amount of drinks equivalent to 30%.

$y = \frac {30 * 20} {100}\\y = 6$

So Tammy drank 6 drinks.

Adding up we have:

$3 + 4 + 5 + 6 = 18$

So: $20-18=2$

Thus, 2 drinks remain in the refrigerator.

Answer:

Option B

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3 years ago

Find the coordinates of the intersection of the medians of △ ????BC given ????(2, 4), ????(−4, 0), and ????(3, −1).

Damm [24]

Answer:

The coordinates of the intersection of the medians of △ABC is (1/3, 1).

Step-by-step explanation:

Consider the vertices of △ABC are A(2, 4), B(−4, 0), and C(3, −1).

Intersection of the medians of a triangle is known as centroid.

Formula for centroid of a triangle is

$Centroid=(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3})$

Using the above formula the centroid of △ABC is

$Centroid=(\dfrac{2-4+3}{3},\dfrac{4+0-1}{3})$

$Centroid=(\dfrac{1}{3},\dfrac{3}{3})$

$Centroid=(\dfrac{1}{3},1)$

Therefore the coordinates of the intersection of the medians of △ABC is (1/3, 1).

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3 years ago