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Paladinen [302]
3 years ago
11

Which types of polygons are the faces of a tetrahedron?

Mathematics
2 answers:
adell [148]3 years ago
6 0

Answer:equilateral triangles

Step-by-step explanation:

Apex

valentina_108 [34]3 years ago
4 0

The answer is the option A, which is: A. Equilateral triangles.

The explanation for this answer is shown below:

By definition, a tetrahedron is a solid in three dimensions that has: six sides, four vertices and four triangular faces. These faces are equilateral triangles, which means that all the sides are equal and all the interior angles measure 60 degrees.

The tetrahedron is also called "triangular pyramid".

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Are (121 over 11),(23 over 3), and (85 over 5) in simplest form?
Sedaia [141]
Answers:

121/11=11

23/3=23/3

85/5=17
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3 years ago
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I do all things in zero sums in my head is that possible to be a zero?
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Yes but it dose depend on the problem
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2 years ago
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Sarah measured her square garden to have an area of 122 ft​ 2​ . How long is one side of Sarah’s garden?
Mashutka [201]

Answer:

30.5 ft squared

Step-by-step explanation:

First, find the area and shape. If it is a square then you want to divide by four since all sides are equal. So divide 122 by 4 and you get the answer.

<h2><u><em>If you need more help or explanation, follow up in the comments below. :P</em></u></h2>
5 0
3 years ago
(-5, 3), (2, 1) what is the slope of the line that passes through the points
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3 0
3 years ago
Weights of cars passing over a bridge part 2<br>​
DaniilM [7]

Step-by-step explanation:

The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.

Given:

mean,  mu = 3550 lbs (hope I read the first five correctly, and it's not a six)

standard deviation, sigma = 870 lbs

weights are normally distributed, and assume large samples.

Probability to be estimated between W1=2800 and W2=4500 lbs.

Solution:

We calculate Z-scores for each of the limits in order to estimate probabilities from tables.

For W1 (lower limit),

Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069

From tables, P(Z<Z1) = 0.194325

For W2 (upper limit):

Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954

From tables, P(Z<Z2) = 0.862573

Therefore probability that weight is between W1 and W2 is

P( W1 < W < W2 )

= P(Z1 < Z < Z2)

= P(Z<Z2) - P(Z<Z1)

= 0.862573 - 0.194325

= 0.668248

= 0.67 (to the hundredth)

6 0
3 years ago
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