You could cross cancel the a and your final answer would be 5/11
Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by
The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same
Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
35a + 25c = 2660
a = 2c
35(2c) + 25c = 2660
70c + 25c = 2660
95c = 2660
c = 2660/95
c = 28 <=== there were 28 children tickets
a = 2c
a = 2(28)
a = 56 <== there were 56 adult tickets
Answer:
29
Step-by-step explanation:
The answer for this question is 1,100