The inverse of p q is:?

-20x-10=30<br> -10x-7y=13

MakcuM [25]

Answer:

$x = - 2\\ 10x + 7y + 13 = 0$

5 0

3 years ago

The total area of the figure to the right is 299 cm². Use this fact to write an equation involving x. Then solve the

Sliva [168]

Answer:

i. $x^{2}$ + 26x + 144 - 299 = 0

ii. x = 5

Step-by-step explanation:

The area of the figure = 299 cm²

The figure has a shape of a rectangle, so that;

Area of rectangle = length x width

length of the figure = 18 + x

width of the figure = 8 + x

Area of the figure = (18 + x) x (8 + x)

299 = (18 + x) x (8 + x)

299 = $x^{2}$ + 26x + 144

$x^{2}$ + 26x + 144 - 299 = 0

$x^{2}$ + 26x - 155 = 0

applying the trigonometric formula;

x = (-b ± $\sqrt{b^{2} - 4ac}$ ) ÷ 2a

= (-26 ± $\sqrt{(26)^{2} - 4(1*-155)}$ ) ÷ 2

= (-26 ± $\sqrt{676 +620}$ ) ÷ 2

= (-26 ± $\sqrt{1296}$ ) ÷ 2

= (-26 ± 36) ÷ 2

x = (-26 + 36) ÷ 2 OR x = (-26 -36) ÷ 2

= 10 ÷ 2 OR -62 ÷ 2

= 5 OR -31

Thus, x = 5

5 0

3 years ago

When simplified completely, the product of a binomial and a binomial is a ______________ monomial.

enyata [817]

Answer:

The answer is Never

8 0

3 years ago

Read 2 more answers

How do I do this!!!!!

Readme [11.4K]

The expected values of the binomial distribution are given as follows:

1. 214.

2. 21.

3. 31.

<h3>What is the binomial probability distribution?</h3>

It is the <u>probability of exactly x successes on n repeated trials, with p probability</u> of a success on each trial.

The expected value of the binomial distribution is:

E(X) = np

For item 1, the parameters are:

p = 3/7, n = 500.

Hence the expected value is:

E(X) = np = 500 x 3/7 = 1500/7 = 214.

For item 2, the parameters are:

p = 0.083, n = 250.

Hence the expected value is:

E(X) = np = 250 x 0.083 = 21.

For item 3, the parameters are:

p = 1/13, n = 400.

Hence the expected value is:

E(X) = np = 400 x 1/13 = 31.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

6 0

2 years ago

An island is 1 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that i

Elanso [62]

Answer:

The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island

Step-by-step explanation:

From the question, we have;

The distance of the island from the shoreline = 1 mile

The distance the person is staying from the point on the shoreline = 15 mile

The rate at which the visitor runs = 6 mph

The rate at which the visitor swims = 2.5 mph

Let 'x' represent the distance the person runs, we have;

The distance to swim = $\sqrt{(15-x)^2+1^2}$

The total time, 't', is given as follows;

$t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}$

The minimum value of 't' is found by differentiating with an online tool, as follows;

$\dfrac{dt}{dx} = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} = \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }$

At the maximum/minimum point, we have;

$\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0$

Simplifying, with a graphing calculator, we get;

-4.72·x² + 142·x - 1,070 = 0

From which we also get x ≈ 15.04 and x ≈ 0.64956

x ≈ 15.04 mile

Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi

$t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89$

Therefore, the distance to run, x ≈ 14.96 mile

6 0

3 years ago