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Mariana [72]
3 years ago
11

How much should Jenna save up to prepare for car repairs and maintenance? (A lump sum emergency fund just for car repair/mainten

ance that may come up in the future)
Mathematics
1 answer:
Veronika [31]3 years ago
8 0

Answer:

$500 per year.

Step-by-step explanation:

Car repairs and maintenance expense is based on the usage of car as well as its present condition. If the car is bought from showroom and is new then it will incur only small amount of expense as its maintenance but if the car is used by various owners then it is likely that maintenance expense can be huge. The average car repair expense is $500 to $800 per year depending on car condition. Since Jenna car is new she should keep the repair estimate to be minimum that is nearly $500 per year or $42 per month.

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Roll 2 six-sided number cubez how many possibilties contain only even numbers
jeka94

Answer:

1/2 of the possibilities are even number

Step-by-step explanation:

Add all of the sides together, you get 12, count on even numbers until you get to 12. 2,4,6,8,10,12. that's six numbers which is half of 12. so half of the possibilities are even.

6 0
3 years ago
A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan
Artyom0805 [142]

Answer:

1) \frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let y represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: \frac{dy}{dt}=rate\:in-rate\:out

The amount coming in is 0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}

The rate going out depends on the concentration of salt in the tank at time t.

If there is y(t) pounds of  salt and there are 100+2t gallons at time t, then the concentration is: \frac{y(t)}{2t+100}

The rate of liquid leaving is is 3gal\min, so rate out is =\frac{3y(t)}{2t+100}

The required differential equation becomes:

\frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) We rewrite to obtain:

\frac{dy}{dt}+\frac{3}{2t+100}y=2.5

We multiply through by the integrating factor: e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }

to get:

(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }

This gives us:

((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }

We integrate both sides with respect to t to get:

(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C

Multiply through by: (50+t)^{-\frac{3}{2}} to get:

y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }

y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}

We apply the initial condition: y(0)=0

0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}

C=-12500\sqrt{2}

The amount of salt in the tank at time t is:

y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}

y(50)=98.23

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})

As t\to \infty, 50+t\to \infty and \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0

This implies that:

\lim_{t \to \infty}y(t)=\infty- 0=\infty

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0
3 years ago
The expression (4c − 3d)(3c + d) is equivalent to:
Umnica [9.8K]

(4c - 3d)(3c + d) =

= 12c² + 4cd - 9cd - 3d² =

= <u>12c² - 5cd - 3d²</u>

4 0
3 years ago
What is the simplified form of (-3x3y2) (5xy-1)
arlik [135]
Simplify the expression
-15x4y3+3x3y2
Hope this helps! :)
~Zain
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3 years ago
F(x) = 2x^2+ 5x + 20<br><br> Find f(-9)
8_murik_8 [283]

Answer:

137

Step-by-step explanation:

2(-9)^2+5(-9)+20

2(81)-45+20

162-45+20

117+20

137

8 0
3 years ago
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