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Nookie1986 [14]
2 years ago
8

A parking lot charges $3 to park a car for the first hour and $2 per hour after that. If you use more than one parking space, th

e second and each subsequent car will be charged 75% of what you pay to park just one car.If you park 3 cars for t hours, which function gives the total parking charge?
A. f(t) = 3(3 + 2(t − 1)) B. f(t) = (3 + 2t) + 0.75 × 2(3 + 2t) C. f(t) = (3 + 2(t − 1)) + 0.75 × 2(3 + 2(t − 1)) D. f(t) = (3 + 2t) + 0.75(3 + 2t) + 0.75 × 0.75(3 + 2t) E. f(t) = (3 + 2(t − 1)) + 0.75(3 + 2(t − 1)) + 0.75 × 0.75(3 + 2(t − 1))
Mathematics
1 answer:
Vikki [24]2 years ago
8 0

Given, a parking lot charges $3 for first hour and $2 per hour after that.

So for t hours, the parking lot charges $3 for the first hour and after first hour there is (t-1) hours left.

So for (t-1) hours it will charge $2 per hour.

The charges for (t-1) hours = $2(t-1).

Total charges for t hours for one car = $(3+2(t-1))

Now for the second car, it will charge 75% of the first car.

So the charges for second car

=$[ (3+2(t-1))(75/100)]

=$0.75(3+2(t-1))

There are 3 cars. That parking charges for the third car is also 75% of the first car.

So for third car the parking charges are same as for the second car.

Total parking charges for 3 cars

= $(3+2(t-1))+(0.75(3+2(t-1))+(0.75(3+2(t-1))

= $(3+2(t-1))+(0.75)(2(3+2(t-1))

We have got the required answer here.

The correct option is option C.

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Step-by-step explanation:

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Initial repair cost = $1200

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HJ is twice JK. J is between H and K. If HJ= 4x and HK=78, find Jk.
ss7ja [257]
You'll find it easier to understand if you illustrate the problem as what is shown in the attached picture. From the illustration and the problem description, two equations can be formulated:

HJ = 2JK
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6 0
2 years ago
If 2x2 + y2 = 17 then evaluate the second derivative of y with respect to x when x = 2 and y = 3. Round your answer to 2 decimal
Vera_Pavlovna [14]

Answer:

y''=-1.26

Step-by-step explanation:

We are given that 2x^2+y^2=17

We have evaluate the second order derivative of y w.r.t. x when x=2 and y=3.

Differentiate w.r.t x

Then , we get

4x+2yy'=0

2x+yy'=0

yy'=-2x

y'=-\frac{2x}{y}

Again differentiate w.r.t.x

Then , we get

2+(y')^2+yy''=0 (u\cdot v)'=u'v+v'u)

2+(y')^2+yy''=0

Using value of y'

yy''=-2-(-\frac{2x}{y})^2

y''=-\frac{2+(-\frac{2x}{y})^2}{y}

Substitute x=2 and y=3

Then, we get y''=-\frac{2+(\frac{4}{3})^2}{3}

y''=-\frac{18+16}{9\times 3}=-\frac{34}{27}

Hence,y''=-1.26

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2 years ago
R+5/mn=p solve for m
AnnZ [28]

Answer:

               \bold{m\ =\ \dfrac5{(p-r)n}}

Step-by-step explanation:

                                             \bold{r+\dfrac5{mn}\ =\ p}\\\\ {}\quad-r\qquad-r\\\\{}\ \ \bold{\dfrac5{mn}\ =\ p-r}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\quad\bold{5\ =\ (p-r)^{_\times}(mn)}\\\\\div(p-r)\quad\div(p-r)\\\\{}\ \ \bold{\dfrac5{p-r}\ =\ mn}\\\\{}\quad \ \div n\quad\ \ \div n\\\\\bold{\dfrac5{(p-r)n}\ =\ m}

If you mean (r+5)/mn then:

\bold{\dfrac{r+5}{mn}\ =\ p}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\ \bold{r+5\ =\ pmn}\\\\\div(pn)\quad\div(pn)\\\\{}\ \ \bold{\dfrac{r+5}{pn}\ =\ m}

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