Find the area of a regular decagon with a side length of 14 and an apothem of 25.

The equation of the line that contains diagonal HM is y = 2/3 x + 7.

nirvana33 [79]

Answer:

The slope of the line that contains diagonal OE will be = -3/2

Step-by-step explanation:

We know the slope-intercept form of the line equation

y = mx+b

Where m is the slope and b is the y-intercept

Given the equation of the line that contains diagonal HM is y = 2/3 x + 7

y = 2/3 x + 7

comparing the equation with the slope-intercept form of the line equation

y = mx+b

Thus, slope = m = 2/3

We know that the diagonals are perpendicular bisectors of each other.

As we have to determine the slope of the line that contains diagonal OE.

As the slope of the line that contains diagonal HM = 2/3

We also know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line.

Therefore, the slope of the line that contains diagonal

OE will be = -1/m = -1/(2/3) = -3/2

Hence, the slope of the line that contains diagonal OE will be = -3/2

3 0

2 years ago

1986gram=................litters

sladkih [1.3K]

Answer:

I think you have asked a wrong question there should be 1986mililitre or _kg

4 0

2 years ago

In how many ways can Kame listen to four CD’s assuming he listens to each CD once?

fiasKO [112]

At most she can listen to her music in 16 different ways.

7 0

2 years ago

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Veronika [31]

The expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Given an integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ .

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

$\int\limits^b_a {f(x)} \, dx$ = $\lim_{n \to \infty}$ ∑f(a+iΔx)Δx ,here Δx=(b-a)/n

$\int\limits^5_1 {x/(2+x^{3}) } \, dx$ =f(x)= $x/2+x^{3}$

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)= $[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}$

$\lim_{n \to \infty}$ ∑f(a+iΔx)Δx=

$\lim_{n \to \infty}$ ∑ $n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n$

=4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$

Hence the expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

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5 0

1 year ago

The following data are the distances between a sample of 20 retail stores and a large distribution center. The distances are in

sesenic [268]

Answer:

Variance = 1,227.27

Standard deviation = 35.03

Step-by-step explanation:

To calculate these, we use the following formulas:

Mean = (sum of the values) / n

Variance = ((Σ(x - mean)^2) / (n - 1)

Standard deviation = Variance^0.5

Where;

n = number of values = 20

x = each value

Therefore, we have:

Sum of the values = 29 + 32 + 36 + 40 + 58 + 67 + 68 + 69 + 76 + 86 + 87 + 95 + 96 + 96 + 99 + 106 + 112 + 127 + 145 + 150 = 1,674

Mean = 1,674 / 20 = 83.70

Variance = ((29-83.70)^2 + (32-83.70)^2 + (36-83.70)^2 + (40-83.70)^2 + (58-83.70)^2 + (67-83.70)^2 + (68-83.70)^2 + (69-83.70)^2 + (76-83.70)^2 + (86-83.70)^2 + (87-83.70)^2 + (95-83.70)^2 + (96-83.70)^2 + (96-83.70)^2 + (99-83.70)^2 + (106-83.70)^2 + (112-83.70)^2 + (127-83.70)^2 + (145-83.70)^2 + (150-83.70)^2) / (20 - 1) = 23,318.20 / 19 = 1,227.27

Standard deviation = 1,227.27^0.5 = 35.03

5 0

2 years ago