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tester [92]
3 years ago
8

Solve the equation A = bh for b. b = Ah b = A/h b = h/A b = h – A Solve the equation A = bh for b. b = Ah b = A/h b = h/A b = h

– A
Mathematics
1 answer:
Leto [7]3 years ago
7 0

Answer:

The answer is "b=\frac{A}{h}".

Step-by-step explanation:

Therefore in this question the "bh" were the multiplied and these values includes the variable which is solving by the isolating of "b".

Let's Divides the both sides for h.

\to \frac{A}{h} = \frac{bh}{h}\\\\\to \frac{A}{h} = b\\\\\to b= \frac{A}{h}

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8x + y = -39<br> 2x - 5y = 27<br> Solve the system by elimination
Ahat [919]

Answer:

(-4, -7)

Step-by-step explanation:

21y=-147

y=-7

2x-5x(-7)=27

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5 0
3 years ago
Maria studied the traffic trends in India. She found that the number of cars on the road increases by 10% each year. If there we
lord [1]

Answer:

b. 8,800,000.

Step-by-step explanation:

In maria study the number of cars on the road increases by 10% each year.

Total number of cars in the first year =80 million cars.

We will add 10% of 80 million to 80 milion to enable us get the number of cars for the second year.

For year 2 we have;

10% of 80,000,000+80,000,000

\frac{10}{100}*80,000,000+80,000,000

8,000,000 + 80,000,000

88,000,000

number of cars in year 2 = 88 million

We will add 10% of 88 million to 88 milion to enable us get the number of cars for the third year.

\frac{10}{100}*88,000,000+88,000,000

8,800,000+88,000,000

96,800,000

the number of cars on the road in year 3 compared to year 2 will just be the increase which is in BOLD. That is 8,800,000

<em>or simply subtract the number of cars in year 2 from that in year 3;</em>

96,800,000-88,000,000 = 8,800,000

8 0
3 years ago
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Answer:

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8 0
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Can someone please help me with this?
olchik [2.2K]
Hi there! The answer is $98

When something becomes 0% more expensive it means that the prices have remained the same as the price last year. Since the price last year was $98, the price this year is also $98.
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36 + 12m

Look at attachment

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