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nikitadnepr [17]

2 years ago

5

How do you figure this out? I am having a hard time. The variables x and y are related to proportionally. When x = 8, y = 20. Fi

nd y when x = 80.
When x = 80, y

1 answer:

kvv77 [185]2 years ago

5 0

Answer:

200

Step-by-step explanation:

(80×20)/8

y is equal to 200

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Production increased from 8,200 units to 13,530 units; the percent increase was what?

mamaluj [8]

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Write the equation for:

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Answer:

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Step-by-step explanation:

Suppose number is x

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Simplifying it further

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What is the product of 64and 35?

kolbaska11 [484]

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2 years ago

The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%

liq [111]

<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by $\frac{100+1.8}{100}$ = $\frac{101.8}{100}$ .

So, the population in the year t can be given by $P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}$

Population in the year 2000 = $3,381,000\textrm{x}(\frac{101.8}{100})^{6}$ = $3,762,979.38$

Population in year 2000 = 3,762,979

Let us assume population doubles by year $y$ .

$2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}$

$log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})$

$y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537$

$y$ ≈ $2033$

∴ By 2033, the population doubles.

4 0

3 years ago