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MrRa [10]
3 years ago
9

John joined a video game club. Newer games cost $5 to rent and older games cost $3 to rent John has $30 to spend each month. How

many of each could John rent?
Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
5 0

Answer:

6 newer games and 10 older games

Step-by-step explanation:

To find how many newer games John can purchase, we divide $30 by $5. If we do the simple math in our head, we will get the answer of 6. Therefore, the amount of newer games John can purchase is 6. Now to find the amount of older games John can buy, we apply the same strategy. We take $30 and we divide it by $3. This will get us the answer of 10. Therefore, the amount of older games John can purchase is 10.

Hope this helps and if it does, don't be afraid to rate my answer as well as maybe give it a "Thanks"? (Or even better a "Brainliest"). And if it’s not correct, I am sorry for wasting your time and good luck finding the correct answer :)

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Nezavi [6.7K]

Answer:

1. 0.5

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Step-by-step explanation:

6 0
3 years ago
What is the (GCF) of the sum 44+40=??
svet-max [94.6K]
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3 years ago
(T 4 Marks + C 1 Marks)
bija089 [108]

Answer:

<h2>length= 40cm</h2><h2>width= 120cm</h2><h2>Step-by-step explanation:</h2>

Area of a rectangle= lenght× width

Dimensions of 1st rectangle: L,W

Dimensions of 2nd rectangle: L+10; W+4

Hence,

LW = 480__________(1)

(L-10)(W+4) = 480_______(2)

LW = LW+4L - 10W - 40

4L = 10W +40

2L = 5W + 20

L = (5/2)W + 10_________(3)

But LW = 480 (from equation one)

So, W = 480/L

substitute as= 480/L into equation (3)

L = (5/2)(480/L)+10

L = 1200/L + 10

L^2 - 10L - 1200 = 0

factorising.

we're have,

(L-40)(L+30) = 0

therefore, since L can't be negative , L = 40

L = 40 cm (length)

Then 480/L = W = 120 cm (width)

<h2>therefore L= 40cm and W= 120cm</h2>
4 0
2 years ago
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Step-by-step explanation:

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I believe the answer is B if I understand the question.
If this answer helped you and was correct you can name me "Brainliest." ty :) 
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