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3 years ago

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The math club is having a car wash to raise funds to attend a state competition. They need to raise a minimum of $325.00. The co

st of supplies for the car wash is $25.00. If they expect 40 cars on the day of the car wash, what is the amount they will have to charge each car to meet the minimum goal?

2 answers:

Svetlanka [38]3 years ago

7 0

Answer:

8.75

Step-by-step explanation:

the formula to use would be:

$325 = 40x - 25

add 25 to both sides of the equation to get:

$350 = 40x

divide both sides by 40 to get:

x = 350/40 = $8.75

if they charge $8.75 per car, they will raise 40 * that for a total of $350.

they would then subtract the $25 for the cost of the supplies and they would be left with $325.

that's their minimum goal.

777dan777 [17]3 years ago

6 0

Answer:

8.75

Step-by-step explanation:

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A builder makes drainpipes that drop 1cm over a horizontal distance of 30cm to prevent clogs. A certain drainpipe needs to cover

Ann [662]

Answer:

700.39cm

Step-by-step explanation:

For a horizontal distance of 30cm, the drop is 1cm, now we need to get the drop for horizontal distance of 700 cm to maintain the same slope

Using the concept of cross multiplication

30cm=1 cm drop

700 cm=x

30x=700

X=700/30=70/3=23 ⅓

Utilizing pythagoras theorem, the length of pipe will be

L=√(23⅓)²+700²

L=700.38878092417 cm

Rounded off, the length is approximately 700.39 cm

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2 years ago

What is the value of this expression if h<br> = 8,j = -1, and k<br> =<br> -12

Rashid [163]

Answer:

12

Step-by-step explanation:

The given expression is :

$\dfrac{j^3k}{h^0}$

We need to find the value of this expression when h = 8, j = -1 and k = -12.

Put all the values in the given expression.

$\dfrac{j^3k}{h^0}\\\\=\dfrac{(-1)^3\times (-12)}{(8)^0}\\\\=\dfrac{-1\times -12}{1}\\\\=12$

So, the value of the given expression is 12.

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2 years ago

A new iPhone 13 is on sale for $999. You get a 10% discount for being a student. But, you are buying the phone at the Apple Stor

soldier1979 [14.2K]

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3 years ago

What is the completely factored form of p4 – 16? (p – 2)(p – 2)(p 2)(p 2) (p – 2)(p – 2)(p – 2)(p – 2) (p minus 2) (p 2) (p squa

n200080 [17]

The completely factored form of the provided polynomial is (p -2) (p+ 2) (p² +4). The option 4 is the correct option.

<h3>What is polynomial?</h3>

Polynomial equations is the expression in which the highest power of the unknown variable is n (n is a real number).

The polynomial equation given in the problem is,

$p^4 - 16$

Let the factor form of the polynomial is f(p). Thus,

$f(p)=p^4 - 16\\f(p)=p^4 - 2^4\\f(p)=(p^2)^2- (2^2)^2$

Using the formula of difference of squares, we get,

$f(p)=(p^2-4)(p^2+4)\\f(p)=(p^2-2^2)(p^2+4)\\f(p)=(p-2)(p+2)(p^2+4)$

Thus, the completely factored form of the provided polynomial is (p -2) (p+ 2) (p² +4). The option 4 is the correct option.

Learn more about polynomial here;

brainly.com/question/24380382

3 0

1 year ago

What is 5.316 - 1.942 btw (show ur work) :)

erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

$\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}$

$\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}$

$\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}$

$\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}$

$\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\$

$\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}$

$\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}$

=3.374

6 0

1 year ago