Answer:
fnjfc ij vcd d GB hhv
Step-by-step explanation:
tb f thg. kufc if fyi uuj
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>
Answer:
5
Step-by-step explanation:
6a = (a*-3) + 45
then:
6a = -3a + 45
6a + 3a = 45
9a = 45
a = 45/9
a = 5
Check:
6*5 = (5*-3) + 45
30 = -15 + 45
Answer: its 1/144
Step-by-step explanation:
Find the common denominator for both unlike fractions.
EX: 1 2
12 8
The lowest common factor is both fractions is 24. Multiply both denominators to get 24.
Do not forget to multiply the numerators with the number you used to multiply the denominators together.
You do not have to multiply each side by the same number.