What is the volume of the cylinder shown below?
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Answer:
Step-by-step explanation:
We are asked to find x, a missing side in a triangle.
This is a right triangle because there is a small square in the corner representing a 90 degree or right angle. Therefore, we can use right triangle trigonometry. The three main functions are:
- sinθ= opposite/hypotenuse
- cosθ= adjacent/hypotenuse
- tanθ= opposite/adjacent
Examine the triangle. We will use angle S, measuring 54 degrees, for theta. Side QR, measuring x, is <u>opposite</u> angle S. Side QS, measuring 2.3, is the <u>hypotenuse</u> because it is opposite the right angle. Since we have the opposite and hypotenuse, we will use sine.
- θ= 54
- opposite= x
- hypotenuse = 2.3
We are solving for x, so we must isolate the variable. It is being divided by 2.3 The inverse operation of division is multiplication, so we multiply both sides by 2.3
Round to the nearest tenth. The 6 in the hundredth place to the right tells us to round the 8 up to a 9.
x is approximately <u>1.9</u>
Since you're only asked the ordered pair of D'', it's much easier just to plot and reflect point D twice than to do that for all four points!
Remember that reflecting points is like putting a mirror at the line of reflection or flipping that point over at that line. The reflected point should be the same distance from the line of reflection as the original point.
1) Reflect D over the x-axis to get D'.
D is at (4,1). Draw a line that is perpendicular to the line of reflection and goes through D. D is as far from the line of reflection as D' should be on its other side (both are on that perpendicular line). Since D is 1 unit above the x-axis, that means D' is 1 unit below at (4, -1). See picture 1.
2) Reflect D' over <span>y=x+1 to get D''.
D' is at (4, -1). Draw </span>y=x+1 and the line perpendicular to it going through D''. D'' is the same distance from the line of reflection on the other side. See picture 2. D'' is at (-2, 5).
Answer: D'' is at (-2, 5)
Remember that reflecting points is like putting a mirror at the line of reflection or flipping that point over at that line. The reflected point should be the same distance from the line of reflection as the original point.
1) Reflect D over the x-axis to get D'.
D is at (4,1). Draw a line that is perpendicular to the line of reflection and goes through D. D is as far from the line of reflection as D' should be on its other side (both are on that perpendicular line). Since D is 1 unit above the x-axis, that means D' is 1 unit below at (4, -1). See picture 1.
2) Reflect D' over <span>y=x+1 to get D''.
D' is at (4, -1). Draw </span>y=x+1 and the line perpendicular to it going through D''. D'' is the same distance from the line of reflection on the other side. See picture 2. D'' is at (-2, 5).
Answer: D'' is at (-2, 5)