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3 years ago

Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 6-inch radius A = sq. in.

1 answer:

8 0

$\bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad 
\begin{cases}
s=length~of\\
\qquad a~side\\
-------\\
s=6
\end{cases}\implies A=\cfrac{6^2\sqrt{3}}{4}\implies A=\cfrac{36\sqrt{3}}{4}
\\\\\\
A=9\sqrt{3}$

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2 years ago

Read 2 more answers

-26a-19-35=-84 . show work!!

Wewaii [24]

Step-by-step explanation:

-26a-19-35=-84

-26a-19+(19)-35=-84+(19)

-26a-35=-65

-26a-35+(35)=-65+(35)

-26a=-30

-26a/26=-30/26

7 0

2 years ago

Please help this hurts

Svetllana [295]

Answer:

6x - 11y = -13 is the answer.

Step-by-step explanation:

Let's plug in the points to see what sticks.

Start with (-4, -1)

1) 11x - 6y = 11(-4) - 6(-1) = -44 + 6 = -38 $\neq$ 13

2) 6x - 11y = 6(-4) - 11(-1) = -24 + 11 = -13

3) 6x - 7y = 6(-4) - 7(-1) = -24 + 7 = -17 $\neq$ 17

4) 6x - 11y = 6(-4) - 11(-1) = -24 + 11 = -13 $\neq$ 13

The only one that fits is #2. Let's try the other point to be sure.

2) 6x - 11y = 6(1.5) - 11(2) = 9 - 22 = -13

3 0

2 years ago

The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Pr

aniked [119]

Answer:

(a) $E(X) = 0.383$

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b) $P(x = 4) = 0.000169$

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Step-by-step explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site?

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

$E(X) = \frac{n \times r}{N}$

$E(X) = \frac{10 \times 8}{209}$

$E(X) = \frac{80}{209}$

$E(X) = 0.383$

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site

The probability is given by

$P(x = 4) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}}$

For the given case, we have

N = 209

n = 10

r = 8

x = 4

$P(x = 4) = \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}}$

$P(x = 4) = \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}}$

$$ P(x = 4) = \frac{70 \times 84944276340}{35216131179263320}$

$P(x = 4) = 0.000169$

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

8 0

2 years ago

In each of the following, a variation relationship is given. Find the requested value of the function in each.

Nadya [2.5K]

Answer:

a.y=6

b.-1/8

c.1/2

d.2800

e.-8

f.7.71

Step-by-step explanation:

6 0

2 years ago