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beks73 [17]
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
Mathematics
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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Encik Basra’s age is four times his child’s age. 4 years ago, Encik Basri’s age is six times his child age. Find the age of Enci
Luden [163]

Answer:

4 + 6 = 10 * 4 =40

Step-by-step explanation:

sana makatulong

4 0
3 years ago
What is the equation of a line , in a slope -intercept form , that passes through (5,-3) and has a slope of 2/3? A. y-3= 2/3(x+5
zmey [24]
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.

y - y1 = m(x - x1)
slope(m) = 2/3
(5,-3)....x1 = 5 and y1 = -3
now we sub...but pay very close attention to ur signs
y - (-3) = 2/3(x - 5) = 
y + 3 = 2/3(x - 5) <=== ur answer in point slope form


8 0
3 years ago
What is the solution to the system of equations graphed below?
pentagon [3]

Answer:

C. (3,-2)

Step-by-step explanation:

5 0
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.

Slope of the line: m= \frac{y_{1} - y_{2}  }{x_{1} - x_{2}  } = \frac{7-3}{-5-2} = \frac{4}{-7}= -\frac{4}{7}

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}

Last step: substitute the slope and y intercept into y=mx+b.

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

That is the answer to this problem.

I hope this helps.

3 0
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}\\

6 0
2 years ago
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