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3 years ago

Solve for y. x y + p = 9 A) x(9-p) B) (p-9)/x C) (9-p)/x

1 answer:

7 0

<span>x y + p = 9
subtract p on both sides
xy + p - p = 9 - p
simplify
xy = 9 - p
divide both sides by x
xy/x = (9-p)/x
simplify
y = (9 - p)/x

Answer
</span><span>C) (9 - p)/x</span>

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If m∠APW = 20°, then the measure of ∠WPZ is

insens350 [35]

Answer:

70°

Step-by-step explanation:

A right angle is 90°. 90°-20°=70°

5 0

2 years ago

Read 2 more answers

Kevin will take 4 math tests this term. All of the tests are worth the same number of points. After taking the first 3 tests, hi

Makovka662 [10]

Answer:

Kevin needs 96 points on his last test to raise his mean test score to 90 points.

Step-by-step explanation:

we know that

The mean score is the total of all scores divided by the total number of tests.

Let

x_1 ----> the score in the first math test

x_2 ----> the score in the second math test

x_3 ----> the score in the third math test

x_4 ----> the score in the fourth math test

we have

After taking the first 3 tests, his mean test score is 88 points

$\frac{x_1+x_2+x_3}{3} =88$

$x_1+x_2+x_3=264$ ----> equation A

How many points does he need on his last test to raise his mean test score to 90 points?

$\frac{x_1+x_2+x_3+x_4}{4} =90$

$x_1+x_2+x_3+x_4=360$ ----> equation B

substitute equation A in equation B

$264 + x_4 = 360$

solve for x_4

$x_4 = 360-264$

$x_4 = 96$

Therefore

Kevin needs 96 points on his last test to raise his mean test score to 90 points.

4 0

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