Answer:
S = {AB, AC, BC}
Step-by-step explanation:
<u>Complete Question</u>
The circle is inscribed in triangle PRT. A circle is inscribed in triangle P R T. Points Q, S, and U of the circle are on the sides of the triangle. Point Q is on side P R, point S is on side R T, and point U is on side P T. The length of R S is 5, the length of P U is 8, and the length of U T is 6. Which statements about the figure are true?
Answer:
(B)TU ≅ TS
(D)The length of line segment PR is 13 units.
Step-by-step explanation:
The diagram of the question is drawn for more understanding,
The theorem applied to this problem is that of tangents. All tangents drawn to a circle from the same point are equal.
Therefore:
|PQ|=|PU|=8 Units
|ST|=|UT| =6 Units
|RS|=|RQ|=5 Units
(b)From the above, TU ≅ TS
(d)Line Segment |PR|=|PQ|+|QR|=8+5=`13 Units
The prime factorisation of 600 is given by

Therefore, a = 3, b = 3, c = 5 and d = 2.
Answer: f(x) = (x + 3)(x – 7)
Step-by-step explanation: Use "standard form" of the function and insert values given: vertex (2,-25) intercept point (7,0)
f(x) = a(x-h)² + k from vertex, h is 2 y is -25 from intercept, x is 7 f(x) is 0
to find a, 0 = a(7-2)² +(-25) 0 = a(7-2)² -25 add 25 to both sides
25 = a(5)² 25 = 25a 25/25 = a 1=a (seems useless but verifies implied "a"coefficient is 1)
f(x) = a(x-h)² + k solve to get the quadratic form
f(x) = (x-2)² -25 (x - 2)² is x² -4x +4
f(x) = x² -4x +4 -25 simplify
f(x) = x² -4x - 21 then factor
f(x) = (x + 3)(x - 7)