What are you looking for? looking for x a or b?
A right triangle's longest side is the hypotenuse
let x=longest, y=middle, and z=shortest
x=y+2
y=2z-1
therefore x=(2z-1)+2=2z+1
find z
z^2+y^2=x^2 by Pythagorean theorem
plug in x and y in terms of z
z^2+(2z-1)^2=(2z+1)^2
z^2+4z^2-4z+1=4z^2+4z+1
subtract the right-hand side's value from the left-hand side's
z^2-8z=0
z(z-8)=0
z=0, 8
z cannot be zero as the sides must have some value to it.
Therefore the shortest side is equal to 8
Answer:
D one real solution
Step-by-step explanation:
x^2 - 8x + 16 = 0
This is in the form
ax^2 +bx + c = 0
so we can use the discriminant to determine the number of solutions
b^2 -4ac
(-8)^2 -4(1)(16)
64 - 64
0
Since the discriminant is zero, there is one real solution.
Answer:
The measure of the angles are 46, 89, and 41 degrees.
Step-by-step explanation:
In order to find this, we need to add all 3 angles together and set equal to 180. Then we can find x.
x + 2x - 3 + x - 5 = 180
4x - 8 = 180
4x = 188
x = 46
Now we can find the value of each angle using the value of x.
Angle 1:
x = 46
Angle 2:
2x - 3
2(46) - 3
92 - 3
89
Angle 3:
x - 5
46 - 5
41
Step-by-step explanation:
6z + 8 = 200
subtract 8 from each sides
6z + 8 - 8 = 200 - 8
simplify --> 6z = 192
6z/6 = 192/6
z = 32
