In circle O, RT and SU are diameters. mArc R V = mArc V U = 64°. Thus, option C is correct.
Given that:
mArc R V = mArc V U,
Angle S O R = 13 x degrees
Angle T O U = 15 x - 8 degrees
<h3>How to calculate the angle TOU ?</h3>
∠SOR = ∠TOU (Vertically opposite angles are equal).
Therefore:
13 x = 15x - 8
Subtracting 13x from both sides
13x - 13x = 15x - 8 - 13x
0 = 15x - 13x - 8
2x - 8 = 0
Adding 8 to both sides:
2x - 8 + 8 = 0 + 8
2x = 8
2x/2 = 8/2
x = 4
∠SOR = 13x
= 13(4)
= 52°
∠TOU = 15x - 8
= 15(4) - 8
= 60 - 8
= 52°
Let a = mArc R V = mArc V U
Therefore:
mArc R V + mArc V U + ∠TOU = 180 (sum of angles on a straight line)
Substituting:
a + a + 52 = 180
2a = 180-52
2a = 128
a = 128/2
a= 64°
mArc R V = mArc V U = 64°
In circle O, RT and SU are diameters. mArc R V = mArc V U = 64°. Thus, option C is correct.
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Answer:yes
Step-by-step explanation: the answer does represent a function because if you were to put points in the line wherever, and connect them by drawing a line vertically, it would not cross two points.
42
let the smallest even number be x, than the four even numbers are x, x+2, x+4, x+6, their sum is 180
x+x+2+x+4+x+6=180
4x+12=180
4x=168
x=42
the smallest even number is 42.
Bar graph display directly the varibales which are the rate and ratio of the numbers to visualize and display the results, these contrasting the different outcomes. On the contrary, histogram is used in grouped frequency parameters. Moreover, as little as the given five parameters or data set this will be ineffective and will result to a bar graph only and basically, the suited option is the aforementioned vertical graph to display the numbers. To expound on the definition of histogram it is used when the frequency is grouped. For example the data set of 1-5, 6-10, 11-15 and 16-20 this now can be used and applied to illustrate histogram because of the number and quantity of the given data.<span>
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Answer:
yes it does, -6x2= -18
Step-by-step explanation:
