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3 years ago

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Please please help me

1 answer:

Yanka [14]3 years ago

6 0

Let A, M, and w represent Angelo's distance per week, Marc's distance per week, and the number of weeks of this regimen, respectively.
.. A = w +7 . . . . . . . Angelo's distance goes up 1 mile/week each week
.. M = 2w +4 . . . . . Marc's distance goes up 2 miles/week each week

a) A = M
.. w +7 = 2w +4
.. 3 = w . . . . . . . . . . subtract w+4

After 3 weeks Angelo and Marc will be running the same distance.

b) A = M = 3+7 = 2*3+4 = 10

That distance will be 10 miles per week.

_____
This problem can be a little confusing because "miles per week" is a rate, and we're talking about the rate of increase of that rate: 1 or 2 miles per week per week. It might be easier to think of the numbers as "weekly distance", rather than "miles per week".

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Solve Using Dirac Deltla/discontinuous forcing

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Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're told that $A(0)=0$ .

For $0\le t\le15$ , the salt flows in at a rate of (1/5 lb/gal)*(5 gal/min) = 1 lb/min. When the regulating mechanism fails, 20 lbs of salt is dumped and no more salt flows for $t>15$ . We can capture this in terms of the unit step function $u(t)$ and Dirac delta function $\delta(t)$ as

$\text{rate in}=u(t)-u(t-15)+20\delta(t-15)$

(in lb/min)

The salt from the mixed solution flows out at a rate of

$\text{rate out}=\left(\dfrac{A(t)\,\mathrm{lb}}{50+(5-5)t\,\mathrm{gal}}\right)\left(5\dfrac{\rm gal}{\rm min}\right)=\dfrac A{10}\dfrac{\rm lb}{\rm min}$

Then the amount of salt in the tank at time $t$ changes according to

$\dfrac{\mathrm dA}{\mathrm dt}=u(t)-u(t-15)+20\delta(t-15)-\dfrac A{10}$

Let $\hat A(s)$ denote the Laplace transform of $A(t)$ , $\hat A(s)=\mathcal L_s\{A(t)\}$ . Take the transform of both sides to get

$s\hat A(s)-A(0)=\dfrac1s-\dfrac{e^{-15s}}s+20e^{-15s}-\dfrac1{10}\hat A(s)$

Solve for $\hat A(s)$ , then take the inverse of both sides.

$\hat A(s)=\dfrac{\frac{10-10e^{-15s}}{s^2}+\frac{200e^{-15s}}s}{10s+1}$

$\implies\boxed{A(t)=10-10e^{-t/10}+\left(30e^{3/2-t/10}-10\right)u(t-15)}$

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4 years ago

Please please help me!!!!!

Tamiku [17]

Your just multiplying 7 and 10, so your answer is gonna be B.

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3 years ago

A mobile phone plan has a base fee of $50 per month. The monthly cost increases by $10 for every gigabyte of data used. Let c re

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110 I’m not sure tho

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3 years ago

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Kendra has 3/4 of a bag of hair ribbons. She wants to share so that each of her friends gets 3/8 of the box. How many friends ca

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3/4 = 6/8 and 3/8 is half of 6/8. If she wanted to give each friend 3/8 of the bag, she could give a portion to only 2 friends.

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4 years ago

Jim sells used drones for 70% of the original manufacturer's purchase price of a new drone. Jim also sells new drones for 15% mo

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The function for the amount saved if they purchase a used drone rather than a new drone from Jim's store is f(x) = 0.45x

He sells used drones at 70% of the price (x).

So, the selling price of a used drone is:

$\mathbf{Used = 70\% \times x}$

Express percentage as decimal

$\mathbf{Used = 0.7 \times x}$

$\mathbf{Used = 0.7x}$

Also, he sells new drones at 15% more of the price (x).

So, the selling price of a new drone is:

$\mathbf{New= (1 + 15\%) \times x}$

Express percentage as decimal

$\mathbf{New= (1 + 0.15) \times x}$

$\mathbf{New= 1.15 \times x}$

$\mathbf{New= 1.15x}$

The amount saved is the difference between the prices of the drones.

So, we have:

$\mathbf{Saved =New - Old}$

This gives

$\mathbf{Saved =1.15x- 0.7x}$

Subtract

$\mathbf{Saved =0.45x}$

Express as a function

$\mathbf{f(x) = 0.45x}$

Hence, the function for the amount saved is f(x) = 0.45x

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