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Evgesh-ka [11]
3 years ago
10

kelsey purchases a coupon book with discounts for her favorite coffee shop: every coupon for the coffee shop offers the same dis

count. The table shows her total savings, y, based on the number of coupons, x, used from the book. what are the slope and Y intercept of the line represented by the points shown in the table?

Mathematics
2 answers:
Afina-wow [57]3 years ago
8 0

Answer:

slope :5, y intercept -20

Step-by-step explanation:

if you look at the chart thats the right awnser

1

1

1

1

1

Aliun [14]3 years ago
3 0

9514 1404 393

Answer:

  b.  slope: 2; y-intercept: -20

Step-by-step explanation:

The y-intercept is the first point in the table, where x=0. That eliminates choices A and D.

The slope is the difference in y divided by the difference in x. Using the first and second points in the table, this is ...

  (-10 -(-20))/(5 -0) = 10/5 = 2

The slope is 2; the y-intercept is -20.

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Step-by-step explanation:

You start with:

16(6)+20(3) ≤ 5

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156 ≤5

Which is wrong.

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Need pre-cal help. Will mark best answer brainliest
OlgaM077 [116]

so, let's keep in mind that

\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.


\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}


we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.


\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1


\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill

5 0
2 years ago
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