Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
Answer:
x = 12
m(QS) = 52°
m(PD) = 152°
Step-by-step explanation:
Recall: Angle formed by two secants outside a circle = ½(the difference of the intercepted arcs)
Thus:
m<R = ½[m(PD) - m(QS)]
50° = ½[(12x + 8) - (4x + 4)] => substitution
Solve for x
Multiply both sides by 2
2*50 = (12x + 8) - (4x + 4)
100 = (12x + 8) - (4x + 4)
100 = 12x + 8 - 4x - 4 (distributive property)
Add like terms
100 = 8x + 4
100 - 4 = 8x
96 = 8x
96/8 = x
12 = x
x = 12
✔️m(QS) = 4x + 4 = 4(12) + 4 = 52°
✔️m(PD) = 12x + 8 = 12(12) + 8 = 152°
I’m thinking 58.45? if that makes sense
