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).
An isosceles triangle has congruent base angles, so if one of the base angles measure 54 degrees, then the other base angle does, too.
And a triangle measures up to 180 degrees.
180 = x + 2(54)
180 = x + 108
x = 72
So the answer is B) 72
Answer:
c = 3.4. I hope this is the answer:)
Answer: The answer is (B) ∠SYD.
Step-by-step explanation: As mentioned in the question, two parallel lines PQ and RS are drawn in the attached figure. The transversal CD cut the lines PQ and RS at the points X and Y respectively.
We are given four angles, out of which one should be chosen which is congruent to ∠CXP.
The angles lying on opposite sides of the transversal and outside the two parallel lines are called alternate exterior angles.
For example, in the figure attached, ∠CXP, ∠SYD and ∠CXQ, ∠RYD are pairs of alternate exterior angles.
Now, the theorem of alternate exterior angles states that if the two lines are parallel having a transversal, then alternate exterior angles are congruent to each other.
Thus, we have
∠CXP ≅ ∠SYD.
So, option (B) is correct.
