Given:
Consider the below figure attached with this question.
In Circle Y, Chords R T and S U intersect.
Arc RS is 106 degrees.
The angle that intercepts arc RS is 94 degrees.
To find:
The measure of arc(TU).
Solution:
If two chords intersect each other insider the circle, then the half of sum of intercepted arcs is equal to the angle on intersection of those arcs.
For the given problem,
![\dfrac{1}{2}[m(arc RS)+m(arc TU)]=94^\circ](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Bm%28arc%20RS%29%2Bm%28arc%20TU%29%5D%3D94%5E%5Ccirc)
![\dfrac{1}{2}[106^\circ+m(arc TU)]=94^\circ](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5B106%5E%5Ccirc%2Bm%28arc%20TU%29%5D%3D94%5E%5Ccirc)
The measure of arc(TU) is 82 degrees.
Therefore, the correct option is A.
Answer:
H
Step-by-step explanation:
The answers simplify to:
F) y=2.5x+5
G) y= -5/2x+5
H) y=5/2x+5
J) y= 5/2x-5
Use rise/run to find slope
5/2=slope
y-intercept= 5
Your answer is H
Answer:
(c) 8x^2 -32x +32, repeated root is x=2.
Step-by-step explanation:
A quadratic with repeated roots will be a multiple of a perfect square trinomial. The form of it will be ...
a(x -b)² = ax² -2abx +ab² = a(x² -2bx +b²)
Dividing by the leading coefficient will leave a monic quadratic whose constant is a (positive) perfect square, and whose linear term has a coefficient that is double the root of the constant.
__
<h3>-x^2 + 18x + 81</h3>
Dividing by the leading coefficient gives ...
x^2 -18x -81 . . . . . a negative constant
__
<h3>3x^2 - 6x + 9</h3>
Dividing by the leading coefficient gives ...
x^2 -2x +3 . . . . . . constant is not a perfect square
__
<h3>8x^2 - 32x + 32</h3>
Dividing by the leading coefficient gives ...
x^2 -4x +4 = (x -2)^2 . . . . . has a repeated root of x=2
__
<h3>25x^2 - 30x - 9</h3>
Dividing by the leading coefficient gives ...
x^2 -1.2x -0.36 . . . . . . a negative constant
__
<h3>x^2 - 14x + 196</h3>
The x-coefficient is not 2 times the root of the constant.
14 = √196 ≠ 2√196
151 feet and 1 inch is equal to 151*12+1 inches, or 1813 inches.
I leaerned this neat trick
you don't need to convert to vertex form
where
y=ax^2+bx+c
the x value of the vertex is -b/2a
the y value is found by inserting the x value of the vertex into the equation
so
1x^2-6x+14
-b/2a=-(-6)/2(1)=6/2=3
plug that in
y=x^2-6x+14
y=3^2-6(3)+14
y=9-18+14
y=5
the vertex is (3,5)
