What object is defined using a directrix and a focus? ...?

Mathematics

2 answers:

katovenus [111]3 years ago

5 0

Answer:

the answer is a parabola

otez555 [7]3 years ago

4 0

Conic sections are defined using a directrix (<span>a fixed line used in describing a curve or surface) and a focus, and those are: parabola, ellipse, and hyperbola. </span>

You might be interested in

I WILL give brainliest promise

ella [17]

Perimeter = 26 Explanation: count the sides of the shape then add them all

8 0

3 years ago

Suppose a 95% confidence interval for μ turns out to be (1,000, 2,100). give a definition of what it means to be "95% confident"

IRISSAK [1]

Answer:

In repeated sampling, 95% of the intervals constructed would contain the population mean.

3 0

3 years ago

What is the relationship?

Studentka2010 [4]

What do you think and what have you previously learned on the subject?

8 0

3 years ago

Find the EXACT value of sin(A−B) if cos A = 3/5 where A is in Quadrant IV and cos B = 12/13 where B is in Quadrant IV. Assume al

MissTica

$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

well, for both angles A and B we're on the IV Quadrant, meaning, the sine is negative, the cosine is positive, likewise, the opposite side is negative and the adjacent side for the angle is positive.

$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$