Answer:
The equation for finding the hypotenuse is:
c^2+c^2=h^2
For example, in the first exercise:
8^2+10^2=164^2
For clearing the equation, find out the root of 164=
12.8
Hypotenuse of the first triangle is 12.8!
Let me help you with the next, if you still dont get it:
8^2+13^2=233^2
Root of 233: 15.2
Hypotenuse of second triangle is 15.2!
X = -6y - 12
4x + 5y =-39
4x + 5y = -39
-4(-6y - 12) + 5y = -39
-4(-6y) + 4(12) + 5y = -39
24y + 48 + 5y = -39
29y + 48 = -39
29y = -87
y = -3
x = -6y - 12
x = -6(-3) - 12
x = 28 - 12
x = 6
(x, y) = (6, -3)
By writing an equation you can easily solve this:
18+ X + 4X - 1 = 42
•we need the answer to 4X-1
1. put the parts that include the X parameter in one side and the integer numbers in one side :
X + 4X = 25
5X= 25
X= 5
2. 4X-1 —> 4x5-1 = 19
The input of each of these functions is always an angle, and as you learned in the previous sections, these angles can take on any real number value. Therefore the sine and cosine function have the same domain, the set of all real numbers, \begin{align*}R\end{align*}
9514 1404 393
Explanation:
Here's one way to go at it.
Draw segments AB and CO. Define angles as follows. (The triangles with sides that are radii are all isosceles, so their base angles are congruent.)
x = angle OAB = angle OBA
y = angle OAC = angle OCA
z = angle OBC = angle OCB
Consider the angles at each of the points A, B, C.
At A, we have ...
angle CAB = x + y
At B, we have ...
angle CBA = x + z
At C, we have ...
angle ACB = y + z
The sum of the angles of triangle ABC is 180°, as is the sum of angles in triangle ABO. This gives ...
x + x + ∠AOB = (x+y) +(x+z) +(y+z)
∠AOB = 2(y+z) = 2∠ACB
This shows ∠AOB = 2×∠C, as required.
