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3 years ago

The black figure is an ellipse, and the black line segment is is major axis. What is the length of the blue line segment?

Mathematics

1 answer:

NeTakaya3 years ago

6 0

Answer:

Step-by-step explanation:

The formula we need to solve this is:

$d_1+d_2=2a$

Where

$d_1$ is the distance from one focus to the vertex

$d_2$ is the second distance from another focus to same vertex

Note: Here, vertex point is the point on top. The common point from which the RED and BLUE line meets the 2 foci

$a$ is half the length of the major axis (which is 9 here since total major axis is given as 18)

Now we can plug into the equation and solve:

$d_1+d_2=2a\\8.5+d_2=2(9)\\8.5+d_2=18\\d_2=18-8.5\\d_2=9.5$

D is the correct answer.

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Car rental company chargers it’s customers an initial fee of 30$ plus daily charge of 25$ to rent a car . Jacob has a budget of

nignag [31]

Answer:

Jacob can rent a car for 6 days with $20 left to spare.

Step-by-step explanation:

200 - 30 = 170

170 / 25 = 6.8

25 * 6 = 150

170 - 150 = 20

6 0

3 years ago

Read 2 more answers

The length of a rectangular kitchen floor is 8 feet longer than w, its width. Which expression represents the area of the floor?

Nookie1986 [14]

Answer:

see below

Step-by-step explanation:

We can write the length as w + 8 and since area is length * width our answer is w * (w + 8) = w² + 8w.

7 0

3 years ago

If sin A = 3/8, find the value of cosec A - sec A.

alexira [117]

Answer:

$\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\$

$\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}$

$\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$