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Mathematics

1 answer:

dolphi86 [110]3 years ago

4 0

Answer:

The correct option is 1.

Step-by-step explanation:

It is given than the area of the triangle ABC is 46cm².

Area of a triangle is

$A=\frac{1}{2}\times base\times height$

Base of the triangle is AB and height of the triangle is CD.

$A=\frac{1}{2}\times AB\times CD$

$A=\frac{1}{2}\times AB\times 8$

$A=4\times AB$

The area of the triangle is 46cm².

$46=4\times AB$

$AB=\frac{46}{4}$

$AB=11.5$

Therefore the length of side AB is 11.5 cm. Option 1 is correct.

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Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar

zavuch27 [327]

Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

$\bf \textit{hyperbolas, vertical traverse axis }
\\\\
\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h, k\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}\\
asymptotes\quad y= k\pm \cfrac{a}{b}(x- h)
\end{cases}\\\\
-------------------------------$

$\bf \begin{cases}
h=2\\
k=-3\\
a=6\\
c=2\sqrt{10}
\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2
\\\\\\
4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2
\\\\\\
\sqrt{4}=b\implies 2=b\\\\
-------------------------------\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1$