Angle Bisector Theorem: In geometry, the angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

Now let's place it according to a general formula.

x/2.25 = 4/3

x= 4/3 * 2.25

x= 3

There you go! now you know the answer and the way to do it! Brainliest pweasee if the answer is correct and helpful!

<h2>૮꒰ ˶• ༝ •˶꒱ა</h2><h2>./づᡕᠵ᠊ᡃ࡚ࠢ࠘ ⸝່ࠡࠣ᠊߯᠆ࠣ࠘ᡁࠣ࠘᠊᠊°.~♡︎ Sara Senpie</h2>

5 0

2 years ago

Question 1<br> Find the Mean of the following set of data values.<br> 46, 57, 68, 63, 49, 52, 61, 68

Zielflug [23.3K]

The mean would be 58!! Basically what you’re doing to get the mean is adding all of the numbers together and dividing by how many you’re adding. In this case, you would add all of the numbers and divide by 8 because you have 8 numbers ! :D

3 0

2 years ago

Suppose that receiving stations X, Y, and Z are located on a coordinate plane at the points (4,5), (-6,-6), and (-14,2

Lilit [14]

Answer:

(-2, -3)

Step-by-step explanation:

A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.

_____

The simultaneous equations for the circles can be solved algebraically.

The epicenter is 10 units from X, so lies on the circle ...

(x -4)^2 +(y -5)^2 = 10^2

x^2 -8x +16 +y^2 -10y +25 = 100

x^2 +y^2 -8x -10y = 59

The epicenter is 5 units from Y, so lies on the circle ...

(x +6)^2 +(y +6)^2 = 5^2

x^2 +12x +36 +y^2 +12y +36 = 25

x^2 +y^2 +12x +12y = -47

The epicenter is 13 units from Z, so lies on the circle ...

(x +14)^2 +(y -2)^2 = 13^2

x^2 +28x +196 +y^2 -4y +4 = 169

x^2 +y^2 +28x -4y = -31

Subtracting the second equation from each of the other two, we get ...

(x^2 +y^2 -8x -10y) -(x^2 +y^2 +12x +12y) = (59) -(-47)

-20x -22y = 106 . . . . eq1 -eq2

(x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)

16x -16y = 16 . . . . . . . .eq3 -eq2

These simultaneous linear equations can be solved a variety of ways. We might use substitution:

x = y+1 . . . . . from eq3 -eq2 divided by 16

10(y +1) +11y = -53 . . . . . from eq1 -eq2 divided by -2

21y = -63 . . . . . . . . . . . . simplify, subtract 10

y = -3

x = y+1 = -2

The epicenter is located at (x, y) = (-2, -3).

8 0

3 years ago