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Lyrx [107]
3 years ago
10

Jakes has $20 to spend on notebooks and pencils the notebooks and pencils. the notebooks cost 3.25 and the pencils cost $.50 wha

t is the maximum number of notebooks he can buy if he buys 8 pencil
Mathematics
1 answer:
Leokris [45]3 years ago
3 0
Hey there. So basically, find out how much the pencils and notebooks cost first.
The notebooks cost = $3.25
The pencils cost = $0.50

Then, think about what you need to figure out in this problem.
Jake has $20. You need to find how many notebooks Jake can buy in maximum after buying 8 pencils.

If Jake buys 8 pencils that costs $0.50 each, he spends $4 on the pencils.

So now, to find out how many notebooks he can buy, do 20 minus 4.
Jake's got $16 left.

If the notebooks cost $3.25 each, we need to find out how many notebooks he can buy by dividing them. So, 16 divided by 3.25 equals 4.923... and so on.

That means, Jake can buy 4 notebooks with his remaining money.
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Answer:

Question 9

(a) The distance, Jalaj walks in one day is 4.4 km

(b) The amount Jalaj raises after walking for 22 km at the end of the 5 days is $8

Question 10

(b) The difference between the largest and smallest areas is 2,400 m²

Step-by-step explanation:

Question 9

(a) The distance Jalaj walks in 5 days = 22 km

Whereby Jalaj walks equal distance every day, we have;

The distance, Jalaj walks in one day = 22 km/5 days = 4.4 km/day

The distance, Jalaj walks in one day = 4.4 km

(b) The amount he raises for every kilometer he walks = $1.60

The amount he raises after walking for 22 km at the end of the 5 days = 5 × $1.60  = $8

Question 10

(b) The given side length of the square = 120 meters to the nearest 10 meters

Therefore;

The maximum dimension for the side length of the square = 120 + 10/2 = 125

The largest possible area of the square, A_l = 125 m × 125 m = 15,625 m²

The minimum dimension for the side length of the square = 120 m - 10 m/2 = 115 m

The smallest possible area of the square, A_s = 115 m × 115 m = 13,225 m².

The difference between the largest and smallest areas, A_l - A_s = 15,625 - 13,225 = 2,400 m².

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