Please help me I’ve been stuck on this all week!:(

Mathematics

1 answer:

Zinaida [17]3 years ago

5 0

Answer:

12x^4 + 10x^3-22x^2 - 15x

Step-by-step explanation:

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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:
$d=\sqrt{(3.5-6.5)^2+(5.5-5.5)^2}
\\=\sqrt{(-3)^2+(0)^2}
\\=\sqrt{9+0}=\sqrt{9}=3$

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4 years ago