How do u solve this algebraically please help

Mathematics

1 answer:

Paraphin [41]3 years ago

4 0

1.
1.
1.Let's solve for x.y=2x−2Step 1: Flip the equation.2x−2=yStep 2: Add 2 to both sides.2x−2+2=y+22x=y+2Step 3: Divide both sides by 2.2x2=y+22x=12y+1

Answer:x=12y+1

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3 years ago

use the information provided to write the standard form equation of each hyperbola vertices:(4,14), (4,-10) foci: (4,15), (4,-11

Evgen [1.6K]

So... if you notice the picture below, based on the given vertices, is a hyperbola with a vertical traverse axis

meaning for the equation, the fraction with the "y" variable is the positive fraction, thus $\bf \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1$

so... what' the point h,k for the center? well, just take a peek at the graph, the center is half-way between both vertices, that's h,k

what's the "a" component or the traverse axis... well, is the distance from one vertex to the center, notice the picture, notice how many units from either vertex to the center, that's "a"

now.. what's "b"? well, "b" comes from the conjugate axis, or the other runnning over the x-axis hmm the smaller one in this case.... well, we dunno what "b" is

however, we know the distance from either focus, to the center of the hyperbola, and that distance "c", is $\bf c=\sqrt{a^2+b^2}$

notice the picture, notice the distance from either focus to the center, that's the distance "c", thus $\bf c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b$

that'd be "b"

bearing in mind that $\bf \cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center= ({{ h}},{{ k}})\\\\ b=\sqrt{c^2-a^2}
\end{cases}$