The answer is C
KLM~STU
From the positions you can find which angles are congruent
K~S
KL~ST
...Etc
Answer:
-9
Step-by-step explanation:
8x-4=6x-10
2x-4=-10
2x=-6
x=-3
-3x3= -9
They give us 2 pieces to the puzzle. Both are positive numbers...x and y.
1.) 1 number is 1 less than twice another number. (x = 2y -1)...and
2.) the sum of their squares is 106. (x^2 + y^2 = 106).
substitute the value for x into the second equation.
(2y-1)^2 + y^2 = 106
(2y-1) (2y-1) + y^2 = 106 (use distributive property)
4y^2 - 2y - 2y + 1 + y^2 = 106 (subtract 106 from both sides)
4y^2 - 2y - 2y + 1 + y^2 - 106 = 106 - 106 (combine like terms)
5y^2 - 4y - 105 = 0 (factor)
(y-5) (5y-21) = 0 (set to 0)
y - 5 = 0
y = 5
substitute the 5 into the equation for y (x = 2(5) - 1)
x = 9 if we square 9, we get 81.
subtracted from 106 we have 25...the square root of 25 is 5.
our answers are 5 and 9.
When you solve that you get 30.825 because you divided 123.3 which was your dimensions times 9 and if you divide that by 4 you get 30.825 . Let me know if that is correct if not I will find new solution!
Answer:
The area of the triangle is 18 square units.
Step-by-step explanation:
First, we determine the lengths of segments AB, BC and AC by Pythagorean Theorem:
AB
![AB = \sqrt{(5-2)^{2}+[6-(-1)]^{2}}](https://tex.z-dn.net/?f=AB%20%3D%20%5Csqrt%7B%285-2%29%5E%7B2%7D%2B%5B6-%28-1%29%5D%5E%7B2%7D%7D)
BC
AC
![AC = \sqrt{(-1-2)^{2}+[4-(-1)]^{2}}](https://tex.z-dn.net/?f=AC%20%3D%20%5Csqrt%7B%28-1-2%29%5E%7B2%7D%2B%5B4-%28-1%29%5D%5E%7B2%7D%7D)
Now we determine the area of the triangle by Heron's formula:
(1)
(2)
Where:
- Area of the triangle.
- Semiparameter.
If we know that , and , then the area of the triangle is:
The area of the triangle is 18 square units.
