There are 12 such shapes. One can start with two squares, adding squares around the edge and rejecting shapes that have been previously seen. Repeating until you have 5 squares results in 12 distinct shapes.
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2 shapes are possible with 3 squares: L, I
5 shapes are possible with 4 squares: L, I, N, T, O
Given problem:
Which value cannot represent the probability of an event occurring?
Values:
0.29
Three-halflity
Probability is the likelihood of an event to occur.
For an event that we are sure of occurring, the probability value is 1, and those that cannot occur have a probability value of 0.
The probability of any event is usually a fraction between 1 and 0.
So we can see that the odd one is the Three-halflity which is clearly greater than 1.
9514 1404 393
Answer:
see attachments
Step-by-step explanation:
The side lengths are found using the distance formula.
d = √((x2-x1)^2 +(y2-y1)^2)
For example, the length of RS in the first triangle is ...
d = √((4-(-1))^2 +(4-(-3))^2) = √(25 +49) = √74 ≈ 8.602
This same computation is used for all of the sides of the triangles. When the same tedious computation is repeated over and over, I like to use a spreadsheet to do it. In the attached spreadsheet the two points used are the one on the same line as the distance, and the one on the line below. Side lengths are shown in the attachment to 3 decimal places.
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In order to classify the triangle as acute, right, or obtuse, we can compare the side lengths to those of a right triangle. For the purposes of computation in a spreadsheet, it is convenient to compute the middle length side of a right triangle that has the same longest and shortest sides as the triangle we have.
If the computed side is longer than the actual middle side, then the triangle we have is obtuse. If it is shorter, the triangle we have is acute. If it is exactly the same, then the triangle we have is a right triangle.
In every case here, the computed middle side is shorter than the actual middle-length side, so all of the triangles are acute triangles. (The one of problem 39 is isosceles, since two sides are the same length. The others are scalene.)
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The triangles are graphed in the second attachment.
problem 38: RST
problem 39: ABC
problem 40: DEF
