Sense the two triangles are congruent you would use CPCTC.
Answer:
Part 1) m∠1 =(1/2)[arc SP+arc QR]
Part 2) 
Part 3) PQ=PR
Part 4) m∠QPT=(1/2)[arc QT-arc QS]
Step-by-step explanation:
Part 1)
we know that
The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.
we have
m∠1 -----> is the inner angle
The arcs that comprise it and its opposite are arc SP and arc QR
so
m∠1 =(1/2)[arc SP+arc QR]
Part 2)
we know that
The <u>Intersecting Secant-Tangent Theorem,</u> states that the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.
so
In this problem we have that

Part 3)
we know that
The <u>Tangent-Tangent Theorem</u> states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments
so
In this problem
PQ=PR
Part 4)
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
In this problem
m∠QPT -----> is the outer angle
The arcs that it encompasses are arc QT and arc QS
therefore
m∠QPT=(1/2)[arc QT-arc QS]
<h3>
Answer: Choice C) 31</h3>
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Explanation:
The recursive rule
f(n+1)=f(n)-3
can be rearranged to
f(n) = f(n+1)+3
after adding 3 to both sides
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Now let's say we plug in n = 3
f(n) = f(n+1)+3
f(3) = f(3+1)+3
f(3) = f(4)+3
f(3) = 22+3
f(3) = 25
Repeat for n = 2
f(n) = f(n+1)+3
f(2) = f(2+1)+3
f(2) = f(3)+3
f(2) = 25+3
f(2) = 28
Each time we keep adding 3 to get the previous term (since the original recursive rule says to subtract 3 to get the next term; we just go backwards of what the instructions say).
Lastly, we can find that f(1) = f(2)+3 = 28+3 = 31 making the answer to be choice C.
Quadratic equation is

WHEN

So, to make this true subtract 7 from each side.
The equation is now
7

- x - 7 = 0
Values for a, b, and c: a = 7, b = -1, c = -7
Plug these variables into the quadratic equation.
Square -1 to get 1 and multiply -4 x 7 x -7 to get 196.

Multiply 2 x -7 to get -14, the denominator, and add 1 to 196.
Since 197 is not a square number, and I don't know if you want to leave it in radical form or not, I will just because it's easier to understand.
Hope this helps!