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Salsk061 [2.6K]
4 years ago
9

NEED HELP!

Mathematics
1 answer:
lana66690 [7]4 years ago
8 0
You know from the y-intercept that the equation will have a +y^2/12 term in it. Then for y=6, you have
.. 6^2/12 -x^2/b = 1
.. 2 = x^2/b
.. b = x^2/2
If your point is (2√3, 6), then this is
.. b = (2√3)^2/2 = 12/2 = 6

Then the hyperbola's equation is
.. y^2/12 -x^2/6 = 1 . . . . . . . . selection D
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What is the sum of the distances of R to Q and P to Q?
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The sum of the distances of R to Q and P to Q is 9 units.

Solution:

Given points are P(–3, 6), Q(3, 6) and R(3, 3).

Distance between two points formula:

   \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Distance from R to Q:

Here x_1=3, y_1=3, x_2=3, y_2=6

Substitute these in the given formula, we get

Distance = \sqrt{(3-3)^2+(6-3)^2}  

               =\sqrt{0+9}

               = 3 units

Distance from R to Q is 3 units.

Distance from P to Q:

Here x_1=-3, y_1=6, x_2=3, y_2=6

Substitute these in the given formula, we get

Distance = \sqrt{(3-(-3))^2+(6-6)^2}  

               =\sqrt{36+0}

               = 6 units

Distance from P to Q is 6 units.

Sum of the distances = 3 + 6 = 9 units

Hence the sum of the distances of R to Q and P to Q is 9 units.

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