Identify the slope and y-intercept of the lins given by the equation y=2x 1.

Find the value of x of the circle and show the work with an equation:

frez [133]

It is given that the measurement of arc AC = 155 . And AB is the diameter, so measurement of arc AB=180 degree . Therefore arc BC = arc AB - arc AC = 180-155=25 . And the radius divides chord CD in two equal parts, so it divides the arc too in two equal parts. Therefore if the measurement of arc CB =25 , then the measurement of arc BD =25 too. Hence x =25 .

5 0

3 years ago

Two lines pass through the origin. The lines have slopes that are opposite. Compare and contrast the lines. What is the slope of

Ira Lisetskai [31]

The compare is hat they both have the same y-intercept, the contrast is that tone line is positive and the other line is negative

6 0

3 years ago

n 2000 the population of a small village was 2,400. With an annual growth rate of approximately 1.68%, compounded continuously,

Nitella [24]

Answer:

The population would be 3358

Step-by-step explanation:

Since, exponential growth function if the growth is compound continuously,

$A = Pe^{rt}$

Where,

P = initial population,

r = growth rate per period,

t = number of periods,

Given,

The population in 2000, P = 2,400,

Growth rate per year = 1.68% = 0.0168,

Number of years from 2000 to 2020, t = 20,

Thus, the population in 2020,

$A=2400 e^{0.0168\times 20}$

$=2400 e^{0.336}$

$= 3358.4136$

$\approx 3358$

6 0

3 years ago

The total monthly profit for a firm is P(x)=6400x−18x^2− (1/3)x^3−40000 dollars, where x is the number of units sold. A maximum

wlad13 [49]

Answer:

Maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

Step-by-step explanation:

We are given the following information: $P(x) = 6400x - 18x^2 - \frac{x^3}{3} - 40000$ , where P(x) is the profit function.

We will use double derivative test to find maximum profit.

Differentiating P(x) with respect to x and equating to zero, we get,

$\displaystyle\frac{d(P(x))}{dx} = 6400 - 36x - x^2$

Equating it to zero we get,

$x^2 + 36x - 6400 = 0$

We use the quadratic formula to find the values of x:

$x = \displaystyle\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$ , where a, b and c are coefficients of $x^2, x^1 , x^0$ respectively.

Putting these value we get x = -100, 64

Now, again differentiating

$\displaystyle\frac{d^2(P(x))}{dx^2} = -36 - 2x$

At x = 64, $\displaystyle\frac{d^2(P(x))}{dx^2} < 0$

Hence, maxima occurs at x = 64.

Therefore, maximum profits are earned when x = 64 that is when 64 units are sold.

Maximum Profit = P(64) = 2,08,490.666667$

6 0

3 years ago

The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The d

kozerog [31]

Answer: 49.85%

Step-by-step explanation:

Given : The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped ( normal distribution ) and has a mean of 61 and a standard deviation of 9.

i.e. $\mu=61$ and $\sigma=9$

To find : The approximate percentage of lightbulb replacement requests numbering between 34 and 61.

i.e. The approximate percentage of lightbulb replacement requests numbering between 34 and $34+3(9)$ .

i.e. i.e. The approximate percentage of lightbulb replacement requests numbering between $\mu$ and $\mu+3(\sigma)$ . (1)

According to the 68-95-99.7 rule, about 99.7% of the population lies within 3 standard deviations from the mean.

i.e. about 49.85% of the population lies below 3 standard deviations from mean and 49.85% of the population lies above 3 standard deviations from mean.

i.e.,The approximate percentage of lightbulb replacement requests numbering between $\mu$ and $\mu+3(\sigma)$ = 49.85%

⇒ The approximate percentage of lightbulb replacement requests numbering between 34 and 61.= 49.85%

4 0

2 years ago