You don't need to understand the construction or why it works. You only need to accept the fact that it does. You can figure out the answers to this question by looking at the picture.
RT is tangent to circle Q -- TRUE. That is the point of the construction.
QT is a radius of circle Q -- TRUE. Q is the center and T is on the circle. A line segment from the center to a point on the circle is a radius.
m∠QSR = 90° -- FALSE. Those points lie on the same line. The measure of the angle is 180°.
QS = QT -- FALSE. S lies inside circle Q, so is closer to the center than T, which lies on the circle. (For some choice of point R, S might lie on the circle, but because this statement is not always true, it must be considered false.)
ΔRTQ is a right triangle -- TRUE. A tangent line is always perpendicular to the radius to the point of tangency. The construction succeeds because RTQ is inscribed in semicircle RTQ (centered at S). Such a triangle is always a right triangle.
Answer:
152 units.
Step-by-step explanation:
(5*8)+(5*8)+(6*8)+(6*4/2)+(6*4/2)
40+40+48+12+12
152
so you're going to need to find the slope. in order to do that use (y2-y1)/(x2-x1)
(7-(-2))/(0-8)
simplify
9/-8
simplify -9/8
now using y=mx+b, we need to find b. so we can plug in the coordinates of (0,7) or (8,-2). let's use (8,-2) because it involves more math.
y=mx+b
-2=(-9/8)(8)+b
multiply (-9/8) by 8 to get -9
-2=-9+b
add 9 to both sides
7=b
so our equation is now
y=-(9/8)x+7
Answer:
<em>f</em> has no critical points.
Step-by-step explanation:
We are given:
A function has critical points whenever its derivative equals 0 or is undefined.
Differentiate the function:
Since this will never be undefined, solve for its zeros:
Hence:
Recall that the value of sine is always between -1 and 1.
Thus, no real solutions exist.
Therefore, <em>f</em> has no critical points.
5/12-1/6 is the correct answer.
