Move 12 to the other side, making it negative
h=31-12
h=19
The equation to represent the area of the triangle would be:
y = 1/2(x²) - (7/2)x
The equation to represent the perimeter of the triangle would be:
y = 3x - 6
The solutions to the system would be (12, 30) or (1, -3). The only viable solution is (12, 30).
Explanation
The area of a triangle is found using the formula
A = 1/2bh
For our triangle, b = x and h = x-7, so we have:
A = 1/2(x)(x-7)
A = 1/2(x²-7x)
A = 1/2(x²) - (7/2)x
We will replace A with y, so we have:
y = 1/2(x²) - (7/2)x
The perimeter of a triangle is found by adding together all sides, so we have:
P = (x-7) + x + (x+1)
Combining like terms we get:
P = 3x - 6
We will replace P with y, so we have:
y = 3x - 6
Since both equations have y isolated on one side, it will be easy to use substitution to solve the system:
3x - 6 = 1/2(x²) - (7/2)x
It's easier to work with whole numbers, so we will multiply everything by 2:
6x - 12 = x² - 7x
We want all of the variables on one side, so we will subtract 6x:
6x - 12 - 6x = x² - 7x - 6x
-12 = x² - 13x
When solving quadratics, we want the equation equal to 0, so we will add 12:
-12+12 = x² - 13x + 12
0 = x² - 13x + 12
This is easy to factor, as there are factors of 12 that sum to -13; -12(-1) = 12 and -12+-1 = -13:
0 = (x-12)(x-1)
Using the zero product property, we know that either x-12=0 or x-1=0; therefore x=12 or x=1.
Putting these back into our equation for perimeter (the simplest one) we have:
y = 3(12)-6 = 36-6 = 30; (12, 30);
y = 3(1) - 6 = 3 - 6 = -3; (1, -3)
We cannot have a negative perimeter, so the only viable solution is (12, 30).
Step-by-step explanation:
Okay I get some of them, your first one - 1 I disagree with. So we can solve with two points by using a equation y2-y1 over x2-x1. So it's 4-8 over - 3-6 with the numerator being - 4 and the denominator being - 9. Any questions on how to solve for slope with two coordinates.
Answer:
Step-by-step explanation:
L.C.M. of 2 and 3=6
Raise to the power 6
