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3 years ago

11

You are planning to volunteer at a zoo that is 60 miles from your home. you can either take stephanie's car, which gets 20 miles

to the gallon

1 answer:

AVprozaik [17]3 years ago

6 0

There is missing information
stephanie's car would require 3 gallons of gasoline to get to the zoo

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Differentiate the function with respect to x<br> f(x)= x^4/ x^2- 4<br> Show work

jeyben [28]

Answer:

Step-by-step explanation:

$f(x)=\frac{u}{v} ,u~ and~ v~ functions~ of~ x\\f'(x)=\frac{v ~u'-u~v' }{v^2} \\f(x)=\frac{x^4}{x^2-4} \\f'(x)=\frac{(x^2-4)*4x^3-x^4(2x)}{(x^2-4)^2} \\=\frac{2x^3(2x^2-8-x^2) }{(x^2-4)^2} \\=\frac{2x^3(x^2-8)}{(x^2-4)^2}$

7 0

2 years ago

I'm sorry but these are the multiple choice answers.

Sunny_sXe [5.5K]

$\bf \qquad \qquad \textit{Future Value of an ordinary annuity}
\\\\
A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]$

$\bf \qquad 
\begin{cases}
A=
\begin{array}{llll}
\textit{original amount}\\
\textit{already compounded}
\end{array}&
\begin{array}{llll}

\end{array}\\
pymnt=\textit{periodic payments}\to &200\\
r=rate\to 7\%\to \frac{7}{100}\to &0.07\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, four quarters}
\end{array}\to &4\\

t=years\to &12
\end{cases}
\\\\\\
A=200\left[ \cfrac{\left( 1+\frac{0.07}{4} \right)^{4\cdot 12}-1}{\frac{0.07}{4}} \right]$

7 0

3 years ago

PLEASE HELP I NEED THIS ASAP

vivado [14]

Answer:

you should've paid attention in class :/

Step-by-step explanation:

8 0

2 years ago

One kilogram is approximately equal to 2.21 pounds. Find the number of pounds in 165 kilograms. Round to the nearest tenth of a

Shtirlitz [24]

Answer: 364.7 pounds

Step-by-step explanation: Multiply the number of kilograms, 165, by the number of pounds per kilogram, 2.21.

165 x 2.21 = 364.65

Round to the nearest tenth.

364.7 pounds in 165 kilograms.

8 0

3 years ago

What is lim x→-3 sqrt x^2-8

vitfil [10]

If the -8 is under the square root, then...

$\displaystyle L = \lim_{x\to -3} \sqrt{x^2-8}\\\\L = \sqrt{(-3)^2-8}\\\\L = \sqrt{9-8}\\\\L = \sqrt{1}\\\\L = 1\\\\$

OR

If the -8 is not under the square root, then...

$\displaystyle L = \lim_{x\to -3} \sqrt{x^2}-8\\\\L = \sqrt{(-3)^2}-8\\\\L = \sqrt{9}-8\\\\L = 3-8\\\\L = -5$

Either way, we replace x with -3 and simplify.

For more information, refer to the direct substitution rule for limits.

4 0

1 year ago