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pshichka [43]
3 years ago
14

Can anyone else not see the answers to questions or is it just me? If so, how do I fix it?

Mathematics
2 answers:
Dimas [21]3 years ago
6 0

Answer:

i cant see it either

Step-by-step explanation:

bixtya [17]3 years ago
5 0

Answer:

Can't see dem either

Step-by-step explanation:

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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]
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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&#10;\qquad &#10;\begin{cases}&#10;center= ({{ h}},{{ k}})\\\\ b=\sqrt{c^2-a^2}&#10;\end{cases}

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3 years ago
Are these ordered pairs in direct variation (1 2) (2 3) (3 4)
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Yes, Because if you draw this into a graph it comes has a straight line going thought the origin
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3 years ago
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