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Svetach [21]
2 years ago
10

Jason went to a carnival where he could go on all rides for a flat free of $30 but he had to pay $2 for each arcade game he play

ed. Jason spent $44. How many arcade games did he play?
a. 22 games
b. 37 games
c. 7 games
d. 15 games
Mathematics
1 answer:
spayn [35]2 years ago
5 0

Answer: C. 7 games

Step-by-step explanation:

2x + 30 = 44  where x is the number of games.

       -30   -30

2x = 14

x= 7

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The amount Cami raised during last year’s charity walk, $45.50, is StartFraction 7 over 10 EndFraction of the amount she raised
barxatty [35]

Answer:

The equation is given by:

n = \frac{7}{10} \times 45.50

Step-by-step explanation:

During last years's charity walk, she raised $45.50.

During this years walk, she raised n, which is 7/10 of this amount. So, the equation is given by:

n = \frac{7}{10} \times 45.50

3 0
3 years ago
Mr . Berg bought 5 number puzzles and 3 word puzzles for his students . The puzzles were $7 each . Mr . Berg used a $60 gift car
DaniilM [7]

Answer:

Mr. Berg got $4 in change.

Step-by-step explanation:

5 0
3 years ago
Quadrilateral CAMP below is a rhombus. The length of PQ is (x + 2) units, and the length of QA is
Phoenix [80]

Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.

<u>Step-by-step explanation:</u>

As given by the statement in the problem,

Q may be the middle point, which cut the diagonal PA into 2 equal halves.

In rhombus, diagonals meet at right angles.

which means that PQ = QA

x+2 = 3x - 14

Grouping the terms, we will get,

3x -x = 14+ 2

2x = 16

dividing by 2 on both sides, we will get,

x = 16/2 = 8

8+2 = 3(8) - 14 = 10 = PQ or QA

3 0
3 years ago
How many books must be chosen from among?
Strike441 [17]
You have to give more explanation
5 0
3 years ago
Given: ABCD is a ∥-gram, BF ⊥ CD , BE ⊥ AD, Prove: △ABE∼△CBF
drek231 [11]

Answer:

Hence proved △ABE∼△CBF.

Step-by-step explanation:

Given,

ABCD is a parallelogram.

BF ⊥ CD    and

BE ⊥ AD

To Prove : △ABE∼△CBF

We have drawn the diagram for your reference.

Proof:

Since ABCD is a parallelogram,

So according to the property of parallelogram opposite angles are equal in measure.

\therefore m\angle A = m\angle B ⇒1

And given that BF ⊥ CD and BE ⊥ AD.

So we can say that;

m\angle F=m\angle E=90\° ⇒2

Now In △ABE and △CBF

∠A = ∠C   (from 1)

∠E = ∠F    (from 2)

So by A.A. similarity postulate;

△ABE∼△CBF

7 0
3 years ago
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