You need to use a ratio of height (H) to shadow length (L) to solve the first problem. It's basically a use of similar triangles, with two perpendicular sides, and with the shadow making the same angle with the vertical.
6 ft = 72 ins, so that rH/L = 72/16 = 9/2 for the player.
So the bleachers are 9/2 x 6 ft = 27 ft.
For the second problem, 9 ft = 108 in, so that the ratio of the actual linear dimensions to the plan's linear dimensions are 9ft/(1/2in) = 2 x 108 = 216.
So the stage will have dimensions 216 times larger than 1.75" by 3".
That would be 31ft 6ins x 54ft.
Live long and prosper.
The length of the diagonal of the canvas is approximately 27 degrees.
The height of the rectangular canvas must reach 18 inches. It must form a 48 degrees angle with the diagonal at the top of the canvas.
<h3>Length of the diagonal Canvas</h3>
Therefore, the length of the diagonal can be found as follows:
Using trigonometric ratio,
- cos ∅ = adjacent / hypotenuse
where
∅ = 48°
adjacent side = Height of the rectangle = 18 inches
hypotenuse = Length of the diagonal
Therefore,
cos 48° = 18 / h
cross multiply
h = 18 / cos 48°
h = 18 / 0.66913060635
h = 26.9005778976
length of the diagonal ≈ 27 inches
learn more on rectangle here: brainly.com/question/26099609?referrer=searchResults
No, a cubic equation can not have three complex roots. This is because it turns twice and one end goes to positive infinity and one end goes to negative infinity. Thus, one of these MUST cross the x-axis at some point, meaning y = 0 and a real root exists.
Yes, a cubic equation can have three real roots if it cuts the x-axis three times.
Answer:
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Step-by-step explanation:
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