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Amiraneli [1.4K]

2 years ago

6

Cory earns d dollars for every hour he works after school. the weekends, his hourly wage is three times the sum of his after sch

ool wage and $2. Write and simplify an expression that represents Cory's weekend wage.

1 answer:

blondinia [14]2 years ago

3 0

Answer:

3d+2

Step-by-step explanation:

Because he is getting 3x the amount of his wages it would be 3xd, simplified to 3d. Then add 2.

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Find the area of the figure

Alex17521 [72]

Answer:

To find area of parallelogram, multiply base and height

Step-by-step explanation:

Base is 6, height is 4

6 x 4 = x

x = 24

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3 years ago

D=.5*a*t^2 - solving for a<img src="https://tex.z-dn.net/?f=d%3D.5%2Aa%2At%5E2" id="TexFormula1" title="d=.5*a*t^2" alt="d=.5*a*

Leviafan [203]

Answer:

$\displaystyle a=\frac{d}{.5*t^2}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$d=.5*a*t^2$

It's required to solve it for a.

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$.5*a*t^2=d$

Divide by $.5*t^2$ :

$\mathbf{\displaystyle a=\frac{d}{.5*t^2}}$

7 0

3 years ago

Some farms use a hay elevator to move bales of hay to the second story of a barn loft. The bottom of the elevator is 9 meters fr

-Dominant- [34]

Answer: The height is 6.07 meters

Step-by-step Explanation: The distance between the elevator and the bottom of the barn is given as 9 meters. Also for the hay elevator to move bales of hay to the second story of the barn lift it makes an angle of elevation of 34 degrees with the ground. With these we can derive a right angled triangle with the reference angle as 34 degrees, the side facing it which is the height or h (opposite) is yet unknown, and the side between the reference angle and the right angle (adjacent) is 9. We shall apply the trigonometric ratio as follows;

Tan 34 = opposite/adjacent

Tan 34 = h/9

0.6745 = h/9

0.6745 x 9 = h

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Therefore the approximate height of the barn to the ground is 6.07 meters

3 0

3 years ago

Read 2 more answers

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago

Let s(t) be the position function of a body moving along a coordinate line; s(t) is measured in feet and t in seconds, where t ≥

Murrr4er [49]

S (0) = (15*0)/(0 + 5) = 0/5 = 0

6 0

3 years ago