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vitfil [10]
3 years ago
10

Let $G$ denote the centroid of triangle $ABC$. If triangle $ABG$ is equilateral with side length 2, then determine the perimeter

of triangle $ABC$. 99 points! for right answer!
Mathematics
1 answer:
belka [17]3 years ago
5 0
Hey there!

You are trying to find the perimeter of triangle ABC. Each side is 2. Since there are 3 sides on a triangle, multiply 2 by 3 to get 6. You do not need the centroid for this problem.

I hope this helps!
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Please help me and give me the answer
snow_tiger [21]

Answer: A) 103.837 rounded to 104.

B)71

Step-by-step explanation:

a) the area of a rectangle is length times width so to find the length we have to divide the area by the width.

b) the perimeter is the length around the shape so the perimeter of a rectangle is all the sides added together. But the length of both sides is the same and so is the width so we had to subtract 98*2 out of 338 then divide that by 2.

5 0
3 years ago
Need the answer now pls
Schach [20]

Answer:

35x-35y

Step by step explanation: I

0.5(20x-50y+36)-0.25(100x+40y-12)

10x-25y+18+25x-10y+3 -------- after we distribute 0.5&-0.25

10x+25x-25y-10y+18+3. ---------we collect the like terms to add them or substract

35x-35y+21 -----------------------------simplified form

6 0
2 years ago
Read 2 more answers
xian and his cousin Kai Both collect stamps, xian has 56 stamps, and Kai have 80 stamps. The boys recently joined different stam
vladimir1956 [14]
In 6 month they will have the same number of stamps. sorry... idk the other part

8 0
3 years ago
1
Kruka [31]

Answer:

24 points

Step-by-step explanation:

We are told timothy usually scores 6 points per quarter of a baektball game.

Now, there are 4 quarters in a basketball game.

Thus, number of points per game = number of points per quarter × number of quarters per game = 6 × 4 = 24 points

6 0
2 years ago
The total scores on the Medical College Admission Test (MCAT) in 2013 follow a Normal distribution with mean 25.3 and standard d
Elden [556K]
In a normal distribution, the median is the same as the mean (25.3). The first quartile is the value of Q_1 such that

\mathbb P(X

You have

\mathbb P(X

For the standard normal distribution, the first quartile is about z\approx-0.6745, and by symmetry the third quartile would be z\approx0.6745. In terms of the MCAT score distribution, these values are

\dfrac{Q_1-25.3}{6.5}=-0.6745\implies Q_1\approx20.9
\dfrac{Q_3-25.3}{6.5}=0.6745\implies Q_3\approx29.7

The interquartile range (IQR) is just the difference between the two quartiles, so the IQR is about 8.8.

The central 80% of the scores have z-scores \pm z such that

\mathbb P(-z

That leaves 10% on either side of this range, which means

\underbrace{\mathbb P(-z

You have

\mathbb P(Z

Converting to MCAT scores,

-1.2816=\dfrac{x_{\text{low}}-25.3}{6.5}\implies x_{\text{low}}\approx17.0
1.2816=\dfrac{x_{\text{high}}-25.3}{6.5}\implies x_{\text{high}}\approx33.6

So the interval that contains the central 80% is (17.0,33.6) (give or take).
7 0
3 years ago
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