Observe the given data distribution table carefully.
The 5th class interval is given as,
The upper limit (UL) and lower limit (LL) of this interval are,
Thus, the upper-class limit of this 5th class is 17.4.
The areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)
<h3>How to determine the total areas?</h3>
<u>The figure 1</u>
In this figure, we have
Length = x + 1
Width = 4
The area is calculated as:
Area = Length * Width
So, we have
Area = 4(x + 1)
<u>The figure 2</u>
In this figure, we have
Length = d + 4
Width = 7
The area is calculated as:
Area = Length * Width
So, we have
Area = 7(d + 4)
<u>The figure 3</u>
In this figure, we have
Length = y + 3
Width = y
The area is calculated as:
Area = Length * Width
So, we have
Area = y(y + 3)
Hence, the areas of the figures are 4(x + 1), 7(d + 4) and y(y + 3)
Read more about areas at:
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Answer:
d 5,15,19
Step-by-step explanation:
The sum of the shortest two sides must be longer than the longest side. That is only the case for selection D.
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If the sum is equal to the longest side, the segment ends will meet, but the "triangle" will look like a line segment. Many authors do not allow such a thing to be called a triangle. Hence, 'b' is not an answer to this question.
(Some authors <em>do</em> allow such a triangle, in which case, there would be two answers here: B and D.)
Answer: 32 and 42
Step-by-step explanation:
5j +12
j= 4
j = 6
5(4) + 12
5 times 4 = 20
20 + 12 = 32
5(6) + 12
5 times 6 = 30
30 + 12 = 42
