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3 years ago

Only do the Evans, PLEASE HELP ME...

Mathematics

2 answers:

Solnce55 [7]3 years ago

8 0

14. t=d/r
15. m=e/c^2
16. c=-p+r

nataly862011 [7]3 years ago

8 0

Answer:

14. t = d/r

15. c=√ 2.718282 m or c=−√ 2.718282 m

16. c=−p+r

17. l = -1.570796w2+a/2w

18. v= 1 /3 bh

19. w=6g−40

Step-by-step explanation:

hope it helps

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2 years ago

Which is the equation of a hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0)? y squared over 40 minus x square

N76 [4]

The equation of the hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0) is $\frac{x^2}{10} + \frac{y^2}{15} = 1$

<h3>How to determine the equation of the hyperbola?</h3>

The given parameters are:

Directrices at x = ±2
Foci at (5, 0) and (−5, 0)

The foci of a hyperbola are represented as:

Foci = (k ± c, h)

The center is:

Center = (h,k)

And the directrix is:

Directrix, x = h ± a²/c

By comparison, we have:

k ± c = ±5

h = 0

h ± a²/c = ±2

Substitute h = 0 in h ± a²/c = ±2

0 ± a²/c = ±2

This gives

a²/c = 2

Multiply both sides by c

a² = 2c

k ± c = ±5 means that:

k ± c = 0 ± 5

By comparison, we have:

k = 0 and c = 5

Substitute c = 5 in a² = 2c

a² = 2 * 5

a² = 10

Next, we calculate b using:

b² = c² - a²

This gives

b² = 5² - 10

Evaluate

b² = 15

The hyperbola is represented as:

$\frac{(x - k)^2}{a^2} + \frac{(y - h)^2}{b^2} = 1$

So, we have:

$\frac{(x - 0)^2}{10} + \frac{(y - 0)^2}{15} = 1$

Evaluate

$\frac{x^2}{10} + \frac{y^2}{15} = 1$

Hence, the equation of the hyperbola is $\frac{x^2}{10} + \frac{y^2}{15} = 1$