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3 years ago

15

at the car wash fundraiser jenna's team washed 14 cars in 45 minutes. if they keep up at that rate, how many cars can the team w

ash in 2.5 hours? how did you get the answer

1 answer:

s2008m [1.1K]3 years ago

4 0

You can write it as a proportion and then solve the proportion.
(2.5 hours is 150 minutes)
14 cars x cars
--------------- = -------------------
45 minutes 150 minutes

Then solve.

x=46.6666666
So that is about 47 cars

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4 + 6 = 10 * 4 =40

Step-by-step explanation:

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Ur asking for slope intercept form, but ur answer choices are in point slope form....so I am gonna find the answer in point slope form.

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3 years ago

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3 years ago

Find the equation of the line that passes through the points (-5,7) and (2,3)

Olegator [25]

Answer:

$y=-\frac{4}{7}x+4\frac{1}{7}$

Step-by-step explanation:

So, in order to solve this problem, I started off by drawing it out. On my graph that I have attached below, I first started out by locating the points (-5,7) and (2,3). Now, this is an optional step, but I highly encourage practicing your graphing skills by solving this problem on graph paper as well. Next, I connected the two points that I just graphed. This is the line that passes through (-5,7) and (2,3).

Now, here is where the actual solving starts. If you haven't already been taught this yet, I will introduce it to you now. I am going to find the equation of this line by filling in what I know in the equation y=mx+b, where m= the slope of the line, and b= y intercept.

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If you haven't been taught how to find the slope of a line I recommend you find out.

Substitute the slope into the equation.

$y=-\frac{4}{7} x+b\\$

Now, we will solve for the 'b,' or y intercept.

We already have x and y values to use: (-5,7) or (2,3). I'll use x=2 and y=3 to solve for the y intercept.

$y=-\frac{4}{7} x+b\\\\3=-\frac{4}{7} *2+b\\\\3=-\frac{8}{7} +b\\b=3+\frac{8}{7} \\b=\frac{21}{7}+\frac{8}{7}=\frac{29}{7} =4\frac{1}{7} \\b=4\frac{1}{7}$

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That is the answer to this problem.

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3 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

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2 years ago