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Zielflug [23.3K]
3 years ago
10

What are vertices of the conic 16x² - 25y² = 400 ?

Mathematics
1 answer:
garri49 [273]3 years ago
5 0

Answer:

(-5, 0) and (5, 0)

Step-by-step explanation:

This equation fits the form for a hyperbola with x-intercepts.  The standard form for such an equation is

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

To get the equation in the question into this standard form, divide each term by 400.

\frac{16x^2}{400}-\frac{25y^2}{400}=\frac{400}{400}\\\frac{x^2}{25}-\frac{y^2}{16}=1

To find the x-intercepts, make <em>y</em> = 0.

\frac{x^2}{25}=1\\x^2=25\\x=\pm 5

The vertices are located at the points (-5, 0) and (5, 0).

<em>Note: </em>There are no <em>y</em>-intercepts; making <em>x</em> = 0 produces no real solutions for <em>y</em>.

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Step-by-step explanation:

Since we are informed that 15 litres of paint covers an area of 250 m2, the the area that 1 liter of paint will cover will be:

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identify which quadrant contains the terminal side of an angle in standard position with each given degree measure with 75 degre
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Answer:

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Step-by-step explanation:

We have the quadratic function:

\displaystyle y=0.3(x-4)^2-2.5

And we want to determine its axis of symmetry.

Notice that this is in vertex form:

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