Find the value of (3/4)²
2 answers:
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Range is set of all y-values. To find a range of graphed function, we need to know that range starts from the minimum value of graph to maximum value. That's because the minimum value is the least value that you can get by substituting the domain and the maximum value is the largest value that you can get by substituting the domain as well.
Now we don't talk about domain here, we talk about range. See the attachment! You are supposed to focus on y-axis, plane or vertical line. See how the minimum value of function is the negative value while the maximum value is positive.
That means any ranges that don't contain the negative values are cleared out. (Hence A and C choices are wrong.)
Next, range being set of all real numbers mean that graphed functions don't have maximum value or minimum value (We can say that both max and min are infinite.)
Take a look at line graph as an example of range being set of all real numbers, or cubic function.
Answer/Conclusion
- The range exists from negative value which is -9 to the maximum value which is 5.
- That means the range is -9<=y<=5
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]