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Eva8 [605]
3 years ago
7

What is the solution to the equation below?

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
5 0
The answer would be c I hope it helps
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Consider the equations of the two lines below:
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Neither, they have a solution

Step-by-step explanation:

Use a sheet of graphing paper and graph the equation using the y-intercept and slope.

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Last month a Carpenter made five tables this month he made 80% more than that how many tables do the carpenter make this month
Nataliya [291]

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URGENT PLEASE HELP PLEASE
mafiozo [28]

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7 0
3 years ago
Given △ABC where A(2, 3), B(5, 8), C(8, 3), RS is the midsegment parallel to AC, ST is the midsegment parallel to AB, and RT is
soldier1979 [14.2K]
Since RS is a midsegment parallel to AC, that means R is the midpoint of AB and S is the midpoint of BC.  The midpoint formula is:
(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}).  Using the coordinates of A and B, we have:
R=(\frac{5+2}{2},\frac{8+3}{2})
\\R=(\frac{7}{2},\frac{11}{2})
\\R=(3.5, 5.5).  Similarly, S is the midpoint of BC:
S=(\frac{5+8}{2},\frac{8+3}{2})
\\S=(\frac{13}{2},\frac{11}{2})
\\S=(6.5, 5.5).
Since ST is a midsegment parallel to AB, then T must be a midpoint of AC:
T=(\frac{2+8}{2},\frac{3+3}{2})
\\T=(\frac{10}{2},\frac{6}{2})
\\T=(5,3).
Now that we have the coordinates of each point we can find the length of each segment using the distance formula:
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
For ST:
d=\sqrt{(6.5-5)^2+(5.5-3)^2}
\\=\sqrt{(1.5)^2+(2.5)^2}
\\=\sqrt{2.25+6.25

\\=\sqrt{8.5}=2.9 \neq 4
For RT:
d=\sqrt{(3.5-5)^2+(5.5-3)^2}
\\=\sqrt{(-1.5)^2+(2.5)^2}
\\=\sqrt{2.25+6.25}
\\=\sqrt{8.5}=2.9 \neq 5
For RS:
d=\sqrt{(3.5-6.5)^2+(5.5-5.5)^2}
\\=\sqrt{(-3)^2+(0)^2}
\\=\sqrt{9+0}=\sqrt{9}=3
6 0
3 years ago
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