ABC is an isosceles triangle so AB = BC
Plug in
x + 4 = 3x - 8
-2x = - 12
x = 6
AC = x = 6
Answer
6 units
Answer: 
Step-by-step explanation:
The perimeter of a rectangle can be calcualated with this formula:
Where "l" is the lenght and "h" is the height.
The area of a rectangle can be found with this formula:
Where "l" is the lenght and "h" is the height.
In this case we know that:
Therfore, we can susbsitute them into and solve for "h" in order to find its value:
Find two number whose sum is 3 and whose product is -40. These are -5 and 8. Then, factorizing, we get:
The positive value is the height. Then:
Since the length is 3 centimeters greater than its height, we get that this is: Then:
Substituting values into , we get that the perimeter is:
(a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2+2ab+2b^2 =The answer
(a + b)^2 = a^2 + 2ab + b^2 => square of sums
(a - b)^2 = a^2 - 2ab + b^2 => square of deference
and of course one of most important ones:
a^2 - b^2 = (a - b)(a + b) => difference of squares
Best Answer: (a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2 + 2ab + 2b^2
a^4 + 4b^4 => i.e. 4a^2b^2 ,
a^4 + 4a^2b^2 + 4b^4 => a^2 + 2ab + b^2 = (a + b)^2, if : a = a^2 , b = 2b^2:
(a^2 + 2b^2)^2 = a^4 + 4a^2b^2 + 4b^4 => We can't add or subtract the value to the expression.
a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 =>
(a^2 + 2b^2)^2 - 4a^2b^2 =>
(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab) =>
(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)
Greetings!
3x - 2y - 1 = 0
y = 5x + 4
3x - 2(5x + 4) - 1 = 0
3x - 10x - 8 - 1 = 0
-7x - 9 = 0
-7x = 9
x = -9/7
y = 5x + 4
y = 5(-9/7) + 4
y = -45/7 + 4
y = -45/7 + 28/7
y = - 17/7
solution is (-9/7, -17/7)
