Triangle Q R S is shown. Line R Q extends through point P. Angle Q S R is 35 degrees. Angle S R Q is 58 degrees. Exterior angle S Q P is x degrees. What is the value of x? [Art not to scale.] x = 23
x = 35
x = 58
x = 93

x = 93

Step-by-step explanation:

Given

<QSR = 35

<SRQ = 58

<SQP = x

Required

Find x

To find the value of x, a triangle needs to be drawn (See attachment).

From the attachment;

<SQP + <SQR = 180 ----------Angle on a straight line (Equation 1)

<SQR + <QSR + <SRQ = 180 ----------- Sum of angles in a s triangle (Equation 2)

Substitute <QSR = 35 and <SRQ = 58 in equation 2

<SQR + 35 + 58 = 180

<SQR + 93 = 180

Subtract 93 from both sides

<SQR + 93 - 93 = 180 - 93

<SQR = 87

SUbstitute 87 for <SQR in Equation 1

<SQP + 87 = 180

Subtract 87 from both sides

<SQP + 87 - 87 = 180 - 87

<SQP = 93

Recall that <SQP = x

So, x = 93

constant acceleration- a

constant velocity- c

no velocity or acceleration- b

explanation:

for body moving with a constant acceleration, the rate of change of displacement changes over time. this is proven by the increasing gradient of the curve (a) against time.

a body that is moving at constant velocity covers equal distances over equal periods of time, thus the straight line with a constant gradient c.

a stationary body   covers no displacement as time elapses. thus the flat line with   zero gradient.

answer: none of these answers are correct only the second half is

step-by-step explanation: