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4 years ago

6

Berlin Germany and el calafate argentina are about the same distance from the equator which of the following is the equation of

the line of best fit

1 answer:

geniusboy [140]4 years ago

7 0

No equation.. this question is no good//

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Please help, I don't understand.. <br><br>Thanks!

olga nikolaevna [1]

Answer:

$109.45

Step-by-step explanation:

On Monday in the morning you worked 4 hours, so plug in the equation.

12.50(4)= $50

On Monday in the afternoon you worked 4 hours and 45 min

12.50(4) = $50

Then do 12.50/60 to get .208 ≈ .21

So 45(.21) = $9.45

50 +50 + 9.45= 109.45

6 0

3 years ago

A tank contains 180 gallons of water and 15 oz of salt. water containing a salt concentration of 17(1+15sint) oz/gal flows into

Stels [109]

Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

$\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}$

and flows out at a rate of

$\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}$

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

$\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))$

Multiply both sides by the integrating factor, $e^{t/180}$ , and rewrite the left side as the derivative of a product.

$e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))$

$\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))$

Integrate both sides with respect to t (integrate the right side by parts):

$\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt$

$\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C$

Solve for A(t) :

$\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}$

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

$\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}$

So,

$\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}$

Recall the angle-sum identity for cosine:

$R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)$

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

$R \cos(\theta) = -\dfrac{66,096,000}{32,401}$

$R \sin(\theta) = \dfrac{367,200}{32,401}$

Recall the Pythagorean identity and definition of tangent,

$\cos^2(x) + \sin^2(x) = 1$

$\tan(x) = \dfrac{\sin(x)}{\cos(x)}$

Then

$R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}$

and

$\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)$

so we can rewrite A(t) as

$\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}$

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

$24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}$

and

$24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}$

which is to say, with amplitude

$2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}$

6 0

2 years ago

Help me please ! Thank you

andreev551 [17]

D I think that’s the answer as dark chocolate takes longer to melt then white

4 0

3 years ago

Read 2 more answers

For the graffiti sweater dodd knit 12 stitches to make 2 inches in width. The sweater is 9 inches wide from the left edge to the

balu736 [363]

Answer:

Step-by-step explanation:

.net 12 stitches to make 2 inches in with the sweater is 9 inches wide from the left edge to the beginning of black smile how many such as wide as that

5 0

3 years ago

Coffee beans worth $ 10.50 $10.50 per pound are to be mixed with coffee beans worth $ 13.00 $13.00 per pound to make 50 50 pound

MAVERICK [17]

Answer: 30 pounds of coffee beans worth $10.50 per pound was used.

20 pounds of coffee beans worth $13.00 per pound was used.

Step-by-step explanation:

Let x represent the number of coffee beans worth $10.50 per pound that should be mixed.

Let y represent the number of coffee beans worth $13.00 per pound that should be mixed.

50 pounds of beans worth $11.50 per pound is to be made. The total cost of the mixture would be

50 × 11.5 = 575

This means that

10.5x + 13y = 575 - - - - - - - 1

The total number of pounds of each type of coffee used is 50. This means that

x + y = 50

Substituting x = 50 - y into equation 1, it becomes.

10.5(50 - y) + 13y = 575

525 - 10.5y + 13y = 575

- 10.5y + 13y = 575 - 525

2.5y = 50

y = 50/2.5 = 20

x = 50 - 20= 30

6 0

4 years ago