Surface area of a triangular prism
2 answers:
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<h3>Answer: roughly 12.6274 meters</h3>
The more accurate value is 12.6274169979696, though it's not fully exact.
Round this however you need to.
The exact distance 32*sin(45) - 10 meters.
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Explanation:
Refer to the diagram below.
I'll use point A in place of point O since the letter 'oh' is very similar looking to the number zero.
Plot point A at the origin (0,0). While at point A, look directly north. Then turn 30 degrees eastward to look at the bearing 030°. Next, move 20 meters along that bearing direction to arrive at point R. Segment AR is 20 meters long.
In the diagram, note how angle RAB is 30 degrees. The side opposite this is BR = m.
We can use the sine ratio to say that
sin(angle) = opposite/hypotenuse
sin(A) = BR/AR
sin(30) = m/20
m = 20*sin(30)
m = 10
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While still at point A, look directly north and turn 135 degrees clockwise (ie toward the east) and move 32 meters along that bearing. You'll arrive at point S as the diagram shows.
Notice how
angle RAB + angle RAC + angle CAS = 30+60+45 = 135
The remaining angle DAS is 180-135 = 45 degrees.
When focusing on triangle DAS, we can say
sin(A) = DS/AS
sin(45) = n/32
n = 32*sin(45)
n = 22.6274169979696
This value is approximate.
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Subtract the values of m and n
n - m = 32*sin(45) - 10 = exact distance
n - m = 22.6274169979696 - 10
n - m = 12.6274169979696
n - m = 12.6274 = approximate distance
Round it however you need to. I'm choosing to round to four decimal places.
So we see that point S is roughly 12.6274 meters east of point R.
If your teacher wants the exact distance, then stick with 32*sin(45)-10.
Answer:
t=0 Sugar = 0 Kg
t=1min Sugar=0.27 Kg
Step-by-step explanation:
Data
Tank = 2640 L (pure water)
Sol=0.09kg Sugar per liter
Vin = Vout = 3L/min
Sugar in the beginning = ?
if beginning is t = 0min Amount of sugar = 0, this is due to the fact that at the moment of entering the tank the content is only water , but if beginning is t= 1min then;