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Serjik [45]
3 years ago
12

The distance traveled in a car varies directly as the amount of gasoline used. If a car travels 520 miles and uses 26 gallons of

gas, find the constant of variation or miles per gallon.
A) 18 mpg
B) 20 mpg
C) 22 mpg
D) 24 mpg
Mathematics
2 answers:
svetlana [45]3 years ago
7 0

Answer:

B

Step-by-step explanation:

givi [52]3 years ago
3 0

Answer: B) 20 mpg

Step-by-step explanation:

Taking into account that the amount of gasoline used is directly variable to the distance traveled in a car, to find the constant of variation in miles per gallon you need to have an end value that will give you how many miles can be travelled with one gallon. The information we have tells us that 520 miles use up 26 gallons, so 520 miles per 26 gallons. To find how many miles are done with only one gallon divide the 520 miles by the 26 gallons. 520/26 =20miles per gallon 20mpg is the constant of variation, which means that the car uses one gallon in 20 miles of travelling, and will use x gallons in 20 times x miles.

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Given the angles in the diagram, what is the angle measurement of the missing angle?
Eva8 [605]

Answer:

150*

Step-by-step explanation:

Using the formula (n -2) x 180* we find that there is 4 sides (n) and in put it and simplify to make (2)x180*. From there we calculate there is 360* in this polygon, and subtract the degrees already provided.  360-210=150

5 0
3 years ago
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Help me on this question. would be nice if you could also graph it as well​
Gre4nikov [31]
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3 years ago
Each investment matures in 3 years. The interest compounds annually.
Sveta_85 [38]

bearing in mind that 4¾ is simply 4.75.

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$600\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=600\left(1+\frac{0.05}{1}\right)^{1\cdot 3}\implies A=600(1.05)^3\implies A=694.575 \\\\[-0.35em] ~\dotfill

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$750\\ r=rate\to 4.75\%\to \frac{4.75}{100}\dotfill &0.0475\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases} \\\\\\ A=750\left(1+\frac{0.0475}{1}\right)^{1\cdot 3}\implies A=750(1.0475)^3\implies A\approx 862.032

well, the interest for each is simply A - P

695.575 - 600 = 95.575.

862.032 - 750 = 112.032.

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3 years ago
Please help me (math)<br><br> Is it A and B or just A? <br><br> Im 99% sure it is not C :)
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I’m pretty sure it’s a
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I will give brainly :) NO LINKS
user100 [1]
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:) fun little riddle
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