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dsp73
3 years ago
15

A delivery truck driver charges a fixed base price of $6 for 2 miles. After 2 miles, he charges an additional $2 for every mile.

After 6 miles, he charges an additional $4 for every mile. Describe the cost of the delivery truck between 1 mile and 2 miles.

Mathematics
2 answers:
Katena32 [7]3 years ago
4 0

The answer is

A.) The cost of the delivery truck between 1 mile and 2 miles is constant.

meriva3 years ago
3 0
The fixed price for up to 2 miles it is $6 if you read a graph of this it is a straight line. It is a constant hence it being a FIXED price
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What is the factored form of 8x^2+12x
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Answer:

4x(2x+3)

Step-by-step explanation:

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kakasveta [241]
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Read 2 more answers
The cost of a book was decreased from 60 to 50 by what percent the price decrased with solution​
Artist 52 [7]

Answer:

The price decreased by 16 1/3%, or approximately 16.3%

Step-by-step explanation:

Use the formula for percent change. If the answer is negative, it is a percent decrease. if the answer is positive, it is a percent increase.

percent change = (new price - old price)/(old price) * 100%

In this case, use:

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percent change = (new price - old price)/(old price) * 100%

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7 0
3 years ago
If angle 1 = 6n + 1 and angle 4 = 4n + 19, what is the angle 2?
kobusy [5.1K]
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6n + 1 = 4n + 19

Subtract 1 and 4n from both sides to get variables on one side and constants on the other. 

6n - 4n = 19 - 1

Combine like terms and solve for n.

2n = 18
n = 18 / 2
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Angles 1 and 2 are congruent as labeled in the graph. 

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Plug in for 9 for n

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Option C is your answer. 


7 0
3 years ago
You invest $300 at 4% interest, compounded every year. What will your balance be after 5 years? (Remember, the formula is A = P(
Stells [14]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill\\ P=\textit{original amount deposited}\to &\$300\\ r=rate\to 4\%\to \frac{4}{100}\to &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{every year, once} \end{array}\to &1\\ t=years\to &5 \end{cases} \\\\\\ A=300\left(1+\frac{0.04}{1}\right)^{1\cdot 5}\implies A=300(1.04)^5\implies A\approx 364.996

3 0
3 years ago
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