Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.
g(x)=-2x+12 (from given table, find slope and y-intercept)
We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.
To show that this is the case,
substitute g(x) into the circle
x^2+(-2x+12)^2=121
x^2+4x^2-2*2*12x+144-121=0
5x^2-48x+23=0
Solve using the quadratic formula,
x=(48 ± √ (48^2-4*5*23) )/10
=0.5058 or 9.0942
So both solutions are real and both have positive x-values.
Answer:
subtract 2.4 from both sides, simplify to -3x=16.8, divide by -3, simplify to x=-5.6.
Step-by-step explanation:
-3x + 2.4 = 19.2
v <u>-2.4</u> <u>-2.4</u>
v 0 16.8
<u>-3x</u>=<u>16.8</u>
-3 -3
x=-5.6
hope this helps,
<h3>Noelle</h3>
<u>Corrected Question</u>
Quadrilateral QRST is inscribed in circle W as shown below. The measure of ∠QRS is 12 degrees less than three times the measure of ∠QTS and
mRQT=mRST .
(a)Determine the measure of Angle QTS .
(b)What is the common measure of angles RQT and RST ?
Answer:
(a)48 degrees
(b)90 degrees
Step-by-step explanation:
Theorem: Opposite angles of a cyclic quadrilateral are supplementary.
(a)
Let the measure of ∠QTS=x
Therefore: m∠QRS=3x-12
∠QTS and ∠QRS are opposite angles of a cyclic quadrilateral. By the theorem above:
x+3x-12=180
4x=180+12
4x=192
x=48 degrees
The measure of angle QTS is 48 degrees.
(b)Since mRQT=mRST
mRQT and mRST are opposite angles of a cyclic quadrilateral
Therefore:
mRQT+mRST=180 degrees
2mRQT=180
mRQT=90 degrees
Therefore, the common measure of RQT and RST is 90 degrees.
Using the line of best fit drawn on the graph, I can estimate from the graph that the possible bank balance is $160.