With geometry. 20 points, and another question worth 98 on my profile!

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 Intersecting Secant-Tangent Theorem, 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 Tangent-Tangent Theorem  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]

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!