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VARVARA [1.3K]
3 years ago
9

​Joe's annual income has been increasing each year by the same dollar amount. The first year his income was ​$17 comma 90017,900

​, and the 44th year his income was ​$20 comma 30020,300. In which year was his income $ 30 comma 700 question mark
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

In 17th year, his income was $30,700.

Step-by-step explanation:

It is given that the income has been increasing each year by the same dollar amount. It means it is linear function.

Income in first year = $17,900

Income in 4th year = $20,300

Let y be the income at x year.

It means the line passes through the point (1,17900) and (4,20300).

If a line passes through two points (x_1,y_1) and (x_2,y_2), then the equation of line is

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

The equation of line is

y-17900=\frac{20300-17900}{4-1}(x-1)

y-17900=\frac{2400}{3}(x-1)

y-17900=800(x-1)

y-17900=800x-800

Add 17900 on both sides.

y=800x-800+17900

y=800x+17100

The income equation is y=800x+17100.

Substitute y=30,700 in the above equation.

30700=800x+17100

Subtract 17100 from both sides.

30700-17100=800x

13600=800x

Divide both sides by 800.

\frac{13600}{800}=x

17=x

Therefore, in 17th year his income was $30,700.

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3 years ago
There is a mountain with 45 bat caves in a row. Every cave has at least 2 bats and there are 490 bats in all. Any 7 caves in a r
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Step-by-step explanation:

The first cave has 7 times more bats than the last cave.  So if the 45th cave has b bats, then the first cave has 7b bats.

There are 77 bats in every row of 7 caves.  So if there are 7b bats in the first cave, then there are 77−7b bats in caves 2 through 7.

Since there are also 77 bats in caves 2 through 8, that means cave #8 must have 7b bats.  Repeating this logic:

#1 = 7b

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#9-#14 = 77−7b

#15 = 7b

#16-21 = 77−7b

#22 = 7b

#23-28 = 77−7b

#29 = 7b

So the first 29 caves have 5(7b) + 4(77−7b) = 308 + 7b bats.

Now we do the same thing from the other end.  If cave #45 has b bats, then caves #39-#44 have 77−b bats.  And since caves #38-44 have 77 bats, then cave #38 has b bats.  Therefore:

#45 = b

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#38 = b

#32-37 = 77−b

#31 = b

So caves 31 through 45 have 3b + 2(77−b) = 154 + b bats.

Adding that to the first 29 caves, plus x number of bats in cave #30:

308 + 7b + x + 154 + b = 462 + 8b + x

We know this equals 490.

490 = 462 + 8b + x

28 = 8b + x

x is a maximum when b is a minimum, which is b = 2.

28 = 8(2) + x

x = 12

There are at most 12 bats in the 30th cave.

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the other equation will be 6.5y+9x=4972 because this represents the total amount of money

plug 2x for y : 6.5(2x)+9x=4972 : x=226

plug 226 in for the first equation : y=2(226) : y=452

8 0
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<h3>​(-3j²k³)²(2j²k)³</h3>

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= 72 j⁽⁴⁺⁶⁾k⁽⁶⁺³⁾ = 72j¹⁰k⁹​

​Answer = 72j¹⁰k⁹

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Step-by-step explanation:

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