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3 years ago

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The Johnson family went out to lunch. How much did the meal cost per person, including sales tax and tip? • The meal cost $46.72

. • The sales tax was 8% of the cost of the meal. • The tip was 15% of the meal and sales tax. • There are 4 people in the Johnson family. A) $14.00 B $14.50 C) $14.95 (D $15.25

1 answer:

dmitriy555 [2]3 years ago

7 0

Answer: c: $14.50

Step-by-step explanation:I just got done doing this question

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Step-by-step explanation:

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A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

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Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

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2 years ago

An object oscillates 4 feet from its minimum height to its maximum height. The object is back at the maximum height every 3 seco

bulgar [2K]

Answer:

$y=2\text{cos}((\frac{2\pi}{3})t)$

Step-by-step explanation:

We have been given that an object oscillates 4 feet from its minimum height to its maximum height. The object is back at the maximum height every 3 seconds. We are asked to find the cosine function that can be used to model the height of the object.

We know that standard form of cosine function is $y = A\cdot \text{cos}(Bt-C)+D$ , where,

|A| = Amplitude,

Period = $\frac{2\pi}{|B|}$ ,

C = Phase shift,

D = Vertical shift.

Since distance between maximum and minimum is 4, therefore, amplitude will be half of it, that is, $A = 2$ .

Since objects gets back to its maximum value in every 3 seconds, therefore, period of the function is 3 seconds. We know that period is given by $\frac{2\pi}{|B|}$ , therefore, we can write $\frac{2\pi}{|B|}=3$ , therefore, $B = \frac{2\pi}{3}$ .

We haven't been given any information about phase and mid-line, we can assume the values of C and D to be zero .

Therefore, our function required function would be $y=2\text{cos}((\frac{2\pi}{3})t)$ .

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