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amid [387]
2 years ago
10

John earned $12 a day for babysitting his siblings after school each weekday (he did not babysit on weekends). He liked to have

a small amount of money to spend on treats at the school canteen that his parents refuse to buy, but he also wanted to build up some savings. John decided to put a fraction of his earnings into a jar straight away. He spent all the rest on canteen treats.
How much money did John earn over a nine week term?

After 9 weeks, John had $378 in his jar. What percentage of his money did he save?
Mathematics
1 answer:
babymother [125]2 years ago
3 0

Answer:

He earns $540

70% of his money went to savings

Step-by-step explanation:

9 weeks x 5 days a week = 45 days

45 days x $12 a day = $540

378 * 100 / 540 =  (378*100) / 540 =  37800/540 = %70

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Evaluate the expression 15g if g = 1/5.
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Answer:

3

Step-by-step explanation:

15g

g=1/5

15×1/5

15/1×1/5=3

6 0
2 years ago
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At soccer practice, for every 5 minutes that Bob runs, e spends 20 minutes practicing dribbling. If Bob keeps the same ratio and
Kitty [74]

5/20=x/36

1/4=x/36

x=9

9 minutes.

8 0
3 years ago
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finlep [7]
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4 0
3 years ago
Add 4 kg 250 g, 2 kg 90 g and 19kg 3 g.
GarryVolchara [31]

Answer:

=25343 grams or 25 kg 343 g.

Step-by-step explanation:

1 kg = 1000 g

∴ 4 kg = 4000 grams

and the 4 kg 250 g would equal = 4250 grams.

∴ 2 kg = 1000 g

2 kg 90 g = 2000 grams + 90 g = 2090 grams

∴ 19 kg = 19000 grams

and 19 kg 3 grams = 19000 grams + 3 grams = 19003 grams

now we will add all of the values together to get the final answer.

4250 grams + 2090 grams + 19003 grams = 25343 grams or 25 kg 343 g.

Hope this helps!!

8 0
3 years ago
Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)
Tems11 [23]

The slope of the tangent line to f at x=-1 is given by the derivative of f at that point:

f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}

Factorize the numerator:

2x^2-2=2(x^2-1)=2(x-1)(x+1)

We have x approaching -1; in particular, this means x\neq-1, so that

\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)

Then

f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4

and the tangent line's equation is

y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2

6 0
2 years ago
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