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VMariaS [17]
3 years ago
5

Vanessa purchased a smart TV for $1,500 on a payment plan. Four months after purchasing the TV, her balance was $1,200. 7 months

after purchasing the TV, her balance was $975. Find an equation that models the balance B(t) after t months. How many months will it take her to pay off the TV loan?
Mathematics
1 answer:
fredd [130]3 years ago
8 0

Answer:

The expression for her balance is "B(t) = -75*t + 1500". It'll take her 20 months to pay off the loan.

Step-by-step explanation:

We are looking for a function that has the following points:

B(0) = 1500

B(4) = 1200

B(7) = 975

Between B(4) and B(7), there was a variation of 3 months and a variation of -225 on her balance, thefore for each month that passed there was a variation of -75. Thefore, the expression for her ballance is:

B(t) = -75*t + 1500

To check if this expression is valid, we will apply the values provided on the problem:

B(0) = -75*0 + 1500 = 1500 (valid)

B(4) = -75*4 + 1500 = -300 + 1500 = 1200 (valid)

B(7) = -75*7 + 1500 = -525 + 1500 = 975 (valid)

When she pays of her loan, B(t) = 0, so we need to solve for t as shown below:

B(t) = -75*t + 1500 = 0

-75*t + 1500 = 0

75*t = 1500

t = 1500 / 75 = 20 months

It'll take her 20 months to pay the loan.

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300\; \rm lb.

Step-by-step explanation:

Let x represent the mass (in pounds) of that 15\% copper alloy required, such that the final mixture would contain 25.5\% copper by mass.

Consider: if x pounds of that 15\% copper alloy is mixed with 700 pounds that 30\% copper alloy, what would be the mass of copper in the mixture?

  • Mass of copper in x pounds of that 15\% copper alloy: (0.15\, x)\; \rm lb.
  • Mass of copper in 700 pounds of that 30\% copper alloy: 700 \times 0.30 = 210\; \rm lb.

Therefore, the mixture would contain (210 + 0.15\, x) \; \rm lb of copper.

The mass of that mixture would be (700 + x)\; \rm lb. The mass fraction of copper in that mixture would be:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%.

This ratio is supposed to be equal to 25.5\%. These two pieces of equations combine to give an equation about x:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%.

\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255.

Simplify and solve for x:

210 + 0.15\, x= 0.255\, (700 + x).

(0.255 - 0.15)\, x= 210 - 0.255 \times 700.

\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300.

Therefore, 300\; \rm lb of that 15\% alloy would be required.

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Answer: the answer is B

Step-by-step explanation:

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