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Zielflug [23.3K]
3 years ago
8

jacque purchased 6 tickets to a hockey game. if the total cost was $225 what is the cost for one ticket

Mathematics
2 answers:
tino4ka555 [31]3 years ago
8 0

Answer:

37.5

Step-by-step explanation:

225÷6........................................

Elena L [17]3 years ago
7 0

Answer: $37.50

Step-by-step explanation: To find the cost for 1 ticket, we need to find the unit price.

Unit price means the cost per unit,

in this case, the cost per ticket.

Since we know that it costs $225 for 6 tickets, to find the

cost for 1 ticket, we need to divide 6 into $225 or 6 into 225.

225 divided by 6 is 37.5.

This means that each ticket costs $37.50.

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1.) You (or your parents) purchase a new car for $19,725.00 plus 4.75% sales tax. The down payment is $2,175.00 and you (or your
Helga [31]

Answer:

$936.94

Step-by-step explanation:

Remember, that because you paid $2,175 up front (down payment), you can borrow $2,175 less from the bank to purchase the car. So first step: Take new car price of $19,725.00 and multiply it by the sales tax percentage ($19,725 * 0.0475) to get $936.94 of tax.

4 0
2 years ago
Please help I need the answer ASAP!
stepladder [879]

Answer:

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Step-by-step explanation:

because u have to find the where the equation is defined. the range is the set  of values that correspond to the domain.

6 0
3 years ago
What is the solution to this equation?
Ugo [173]

Answer:

C. x = 12

Step-by-step explanation:

5(12) + 9 = 69 - 3(12) = 33

18 + 15 = 33

6 0
2 years ago
Read 2 more answers
Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”
Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± \sqrt{1-(y-2)^{2}}

y = ± \sqrt{1-x^{2}}+2

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

 R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

 of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* <em>Lets solve the problem</em>

- The equation of a circle of center (h , k) and radius r is

  (x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒  (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± \sqrt{1-(y-2)^{2}}

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± \sqrt{1-x^{2}}

- Add 2 for both sides

∴ y = ± \sqrt{1-x^{2}}+2

∴ y = h(x)

- In the function x = ± \sqrt{1-(y-2)^{2}}

∵ \sqrt{1-(y-2)^{2}} ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

- In the function y = ± \sqrt{1-x^{2}}+2

∵ \sqrt{1-x^{2}} ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

8 0
3 years ago
Solve negative one equals negative two v plus two-thirds for v.
makkiz [27]
We write it out as an equation:

-1 = -2v + 2/3

Rearrange:
-1 -2/3 = -2v

Multiply by negative to equal positive
1 2/3 = 2v

Make 1 into a fraction
3/3 + 2/3 = 2v
5/3 = 2v
10/6 = 2v
5/6 = v

The answer is: v equals 5/6
3 0
2 years ago
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