The ice skating rink charges an hourly fee for skating and $3 to rent skates for the day. Gillian rented skates and skated for 3 hours and was charged $21. Which equation represents the cost, c(x), of ice skating as a function of x, the number of hours of skating?

Answers

Answer 1
Answer:

Answer:

The equation represents the cost, c(x), of ice skating as a function of x is c(x)=6x+3.

Step-by-step explanation:

Consider the provided information.

Let R be the per hour rate of the using the skating rink.

It is provided that $3 is the rent of skates for the day.

If Gillian rented skated and paid $21 after 3 hours of skating, this can be represented as:

21=3R+3

3R=18

R=(18)/(3)=6

Therefore, the hourly fee for skating is $6.

Now, write the equation represents the cost c(x) of the ice skating as a function of x.

Total cost = Number of hours × Hourly fee + Rent skates for the day.

Replace total cost with c(x), Number of hours with x, Hourly fee with 6 and rent skates of the day with 3.

Thus the required equation is:

c(x)=6x+3

Thus, the equation represents the cost, c(x), of ice skating as a function of x is c(x)=6x+3.

Answer 2
Answer: c(x) = 6x + 3

$6 per hr..and $3 for the skates

check..
c(x) = 6x + 3
he skated for 3 hrs and was charged $21
c(3) = 6(3) + 3
c(3) = 18 + 3
c(3) = 21
.so it is c(x) = 6x + 3

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V=1/3πr^2h solve for h. Please show work.

Answers

The equation V = (1/3) πr²h is solved for h is h = (3V) / (πr²).

Given that:

Equation, V = (1/3) πr²h

In other words, the collection of all feasible values for the parameters that satisfy the specified mathematical equation is the convenient storage of the bunch of equations.

Multiply both sides of the equation by 3 to eliminate the fraction:

3V = πr²h

Divide both sides of the equation by (πr^2) to isolate h:

(3V) / (πr²) = h

More about the solution of the equation link is given below.

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V=1/3*pi*(r^2)*h
first multiply both sides by 3
3V=pi*(r^2)*h
divide by pi
3V/pi=(r^2)*h
divide by r^2
h=3V/(pi*(r^2))

Which sequence is generated by the function f(n + 1) = f(n) – 2 for f(1) = 10?

Answers

Answer: 10, 8, 6 , 4, 2, 0, -2, -4, ...

Explanation: every number of the sequence is the previous less 2 and the first terms is 10.

Answer:

D. 10, 8, 6, 4, 2, ...

Step-by-step explanation:

I got it right on the EDG. unit test!

Two blocks are connected by a very light string passing over a massless and frictionless pulley. The 20.0 N block moves 75.0cm to the right and the 12.0 N block moves 75.0cm downward.Find the total work done on 20.0N block if there is no friction between the table and the 20.0N block.

Answers

The string is assumed to be massless so the tension is the sting above the 12.0 N block has the same magnitude to the horizontal tension pulling to the right of the 20.0 N block. Thus, 
1.22 a = 12.0 - T  (eqn 1)
and for the 20.0 N block: 
2.04 a = T - 20.0 x 0.325 (using µ(k) for the coefficient of friction) 
2.04 a = T - 6.5  (eqn 2) 

[eqn 1] + [eqn 2] → 3.26 a = 5.5 
a = 1.69 m/s² 


Subs a = 1.69 into [eqn 2] → 2.04 x 1.69 = T - 6.5 
T = 9.95 N 

Now want the resultant force acting on the 20.0 N block: 
Resultant force acting on the 20.0 N block = 9.95 - 20.0 x 0.325 = 3.45 N 
Units have to be consistent ... so have to convert 75.0 cm to m: 

75.0 cm = 75.0 cm x [1 m / 100 cm] = 0.750 m 
work done on the 20.0 N block = 3.45 x 0.750 = 2.59 J

Final answer:

The total work done on the 20.0 N block as it moves 75.0 cm to the right, with no friction between the table and the block, is 15 Joules.

Explanation:

In physics, work done by a force is given by the formula Work = Force x Distance x Cos θ, where θ is the angle between the force and the direction of motion. Here, the force is the same as the weight of the 20.0 N block (because weight = mass * gravity, and the mass is the weight divided by gravity, so mass * gravity = weight), and because the block moves horizontally (rightward), θ is 0 degrees. The Cos of 0 degrees is 1.

So, the work done on the 20.0 N block as it moves 75.0 cm (or 0.75 meters, because 1 m = 100 cm) to the right is 20.0 N * 0.75 m * 1 = 15 Joules.

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Which shows 60^2 − 40^2 being evaluated using the difference of squares method?(A)60^2-40^2=(3600+1600)(3600-1600)=10,400,000
(B)6^2-40^2=3600-1600=2,000
(C)60^2-40^2=(60-40)^2=20^2=400
(D)60^2-40^2=(60+40)(60-40)=(100)(20)=2,000

Answers

Answer:

Option D - 60^2-40^2=(60+40)(60-40)=(100)(20)=2000

Step-by-step explanation:

Given : Expression  60^2-40^2

To find : Which shows expression being evaluated using the difference of squares method?

Solution :

The best method to find the difference of squares method is by using identity,

a^2-b^2=(a+b)(a-b)

According to question,

a=60 and b=40

60^2-40^2=(60+40)(60-40)

60^2-40^2=(100)(20)

60^2-40^2=2000

Therefore, Option D is correct.

D. The answer has to be 2,000, eliminating A and C. B doesn't even use the original equation, as well as incorrectly solving the method, so D is the correct answer, even if you don't understand the principles.

A randomly generated list of numbers from 0 to 4 is being used to simulate an event, with the numbers 3 and 4 representing a success. What is the estimated probability of a success?

Answers

Answer:

40%

Step-by-step explanation:

a.p.e.x

2/4=1/2 because 3 and 4 are 2 of four numbers

If ABC ~ DEC, solve for x. The image is not drawn to scale.A.
x = 2
B.
x = 3
C.
x = 4
D.
x = 13

Answers

Answer: A.    x = 2

Step-by-step explanation:

In the given picture we have \triangle{ABC}\sim\triangle{DEC}

Since, we know that the corresponding sides in similar triangles are in proportion.

Therefore, we have

(CD)/(AC)=(CE)/(BC)\n\n\Rightarrow(8+x)/(6)=(19-2x)/(11-x)\n\n\Rightarrow\ (11-x)(8+x)=6(19-2x)\n\n\Rightarrow\ 88 + 3x -x^2=114-12x\n\n\Rightarrow x^2-15x+26=0\n\n\Rightarrow\ (x-13)(x-2)=0\n\n\Rightarrow\ x=13,2

But x can not be 13 because BC=11-13=-2, which is not possible.

Therefore, the value of  x=2.

If\ \Delta ABC\sim\Delta DEC\ then:\n\n(19-2x)/(11-x)=(8+x)/(6)\ where\ x\in(-8;\ 11)\ \ \ |cross\ multiply\n\n(11-x)(8+x)=6(19-2x)\n11(8)+11(x)-x(8)-x(x)=6(19)+6(-2x)\n88+11x-8x-x^2=114-12x\n-x^2+3x+88=114-12x\ \ \ |subtract\ 114\ from\ both\ sides\n-x^2+3x-26=-12x\ \ \ \ \ |add\ 12x\ to\ both\ sides\n-x^2+15x-26=0\ \ \ \ |change\ signs\nx^2-15x+26=0\nx^2-2x-13x+26=0\nx(x-2)-13(x-2)=0\n(x-2)(x-13)=0\iff x-2=0\ or\ x-13=0\n\nx=2\in(-8;\ 11)\ or\ x=13\notin(-8;\ 11)\n\nAnswer:\boxed{\boxed{A.\ x=2}}