A simple answer is that any given trapezoid with height h and length of the parallel lines a and b, is half of a parallelogram with an area of (a+b) x h. Since the trapezoid is half of this, it is h(a+b)/2
we know that
The least
common multiple (LCM) of two or more numbers is the least
number, other than zero, that is a multiple of all the numbers.
In this problem, we are looking for the least common multiple of 
and 
So
Multiply by
therefore
On the 
day after today, she will do both on the same day
the answer is
The perimeter "P" is equal to the length of the base of one triangle multiplied by the "n" number of triangles in the figure plus two times the length of another side. The equation for the perimeter is P = 5n + 14.
We are given triangles. The triangles are arranged in a certain pattern. The length of the base of each triangle is equal to 5 units. The length of the other two sides is 7 units each. We conclude that all the triangles are isosceles. We need to find the relationship between the number of triangles and the perimeter of the figure. Let the perimeter of the figure having "n" number of triangles be represented by the variable "P".
P(1) = 14 + 5(1)
P(2) = 14 + 5(2)
P(3) = 14 + 5(3)
We can see and continue the pattern. The relationship between the perimeter and the number of triangles is given below.
P(n) = 14 + 5n
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Answer:
8 is the correct answer ...
First, you need to distribute the right side of the equation : 8n - 2
Now the equation is : 20 - 7n = 8n - 2
Add 2 to both sides : 22 - 7n = 8n
Add 7n to both sides : 22 = 15n
Divide both sides by 15 : ( I had to round this one) 1.5 = n
