9514 1404 393
Answer:
31.243 units
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you of the relationships between sides and angles in a right triangle. Using the attached figure, it is convenient to find the length of BE as an intermediate step in the solution.
Sin = Opposite/Hypotenuse
sin(30°) = BE/100
BE = 100·sin(30°)
Then ...
Tan = Opposite/Adjacent
tan(58°) = BE/x
x = BE/tan(58°) = 100·sin(30°)/tan(58°)
x ≈ 31.243 . . . . units
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<em>Comment on the figure</em>
The intermediate problem in creating the figure was to locate point D. That was accomplished by locating point C on a line at an angle of 58° CCW from the horizontal, using point B as a center. Then D is the intersection of BC with the x-axis. BE is drawn perpendicular to the x-axis.
Answer:
I think it's 1/2 because it might be right
I think that it might be 62,100 because the 6 is in the ten thousands place, and if you divide it by 10, 60,000/10, then it would be 6000. 60,000 is 10 times more than 6,000. Correct me if I'm wrong.
Step-by-step explanation:
x + y ≤ 6
You'll have to express the linear inequality into an equality statement (in slope-intercept form, y = mx + b) in order to graph. You'll have to use a solid line because of the "≤ " symbol.
y = - x + 6
Next, you'll have to choose two points to use for graphing. It's always safe to start with the y-intercept, (0, 6).
From there, you could use the slope (m = -1) and do the "rise over run" technique to plot your next point. Going up 1, left 1 will get you to (-1, 7); going down 1 right 1 will get you to point (1, 5). Connect your points and create a straight line.
Next, to determine which region to shade, you must choose a convenient test point (that's not on the line) to see whether the given test point will satisfy the inequality statement.
Let's choose the point of origin, (0, 0), and plug its values into the inequality statement:
x + y ≤ 6
0 + 0 ≤ 6
0 ≤ 6 (True statement). This means that you should shade the half-plane region where the test point is located (left side of the plane).
The first screenshot shows how I created the lines by plotting the first three points. The second screenshot shows the actual graph with the shaded region.
Please mark my answer as the Brainliest if you find this explanation helpful :)
You are right!!!!!!!!!!!!!!!
