Answer:
PT = 16.3 units
Step-by-step explanation:
From the picture attached,
From ΔPQR and ΔPST,
QR ║ST [Given]
PQ is a transversal line.
Therefore, ∠PST ≅ ∠PQR [Corresponding angles]
∠P ≅ ∠P [Common angle]
ΔPQR ~ ΔPST [By AA property of similarity of two triangles]
Therefore, by the property of similarity, corresponding sides of the similar triangles will be proportional.
PT = 
PT = 16.3 units
<h3>Answer:</h3>
The two numbers are 17 and 9.
<h3>Explanation:</h3>
<em>System of Equations</em>
Let x and y represent the first and second numbers, respectively.
... x - y = 8 . . . . . . . . . the difference of the two numbers is 8
... x = 2y - 1 . . . . . . . . .the first number is one less than twice the second
<em>Solution</em>
The second equation gives an expression for x that can be substituted into the first equation.
... (2y -1) -y = 8
... y -1 = 8 . . . . . . . collect terms
... y = 9 . . . . . . . . . add 1
... x = 2·9 -1 = 18 -1
... x = 17
The first number is 17; the second number is 9.
Negative is infinite
positive is dependent on whether or not a can = b etc.
The slope is not shown on the problem but whatever the slope is, plug it in the equation below before the x .
Y=__X-2
Hope this helps
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.
