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mihalych1998 [28]
4 years ago
5

Helppppp plsss solve for x

Mathematics
2 answers:
pogonyaev4 years ago
8 0
The two triangles we can see in the diagram are similar, which means that their measurements are proportional. Line MN on the larger triangle MNP is 71.5ft long, and its corresponding line, MA, on the smaller triangle is 71.5-22= 49.5ft long. This means that the scale factor from the bigger to the smaller triangle is 71.5/49.5=13/9.
If line MP on the bigger rectangle is 97.5ft long, then its corresponding line, MB, must be 97.5÷13/9=67.5 ft long.
Therefore, x=67.5 ft
Art [367]4 years ago
6 0
There are two similar triangles.
small one , sides x and 71.5-22=49.5
bigger one, sides 97.5 and 71.5

x/97.5 = 49.5/71.5
x = 67.5
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Step-by-step explanation:

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bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.


\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1

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