-3/5,-6/5,3/4,7/5 i really hope this helps
Answer:
B. 9.9 ft
Step-by-step explanation:
You have a right triangle, so you can use the Pythagorean theorem.
a^2 + b^2 = c^2
15^ + x^2 = 18^2
225 + x^2 = 324
x^2 = 99
x = sqrt(99)
x = 9.9
Answer: B. 9.9 ft
Answer:
59.44 units²
Step-by-step explanation:
The area of each triangular section can be computed from the side length (5 units) and the central angle (72°) as ...
... section area = (1/2)sin(72°)·5² ≈ 12.5·0.951057 ≈ 11.8882 . . . units²
Then 5 such sections will have an area of ...
... pentagon area = 5·11.8882 units² = 59.4410 units²
_____
<em>Comment on central angle</em>
The central angles total 360°. There are 5 equal sections, so the central angle for each of them is 360°/5 = 72°.
<em>Comment on area formula</em>
It can be occasionally handy to know that the area of a triangle can be computed from the lengths of adjacent sides and the angle between them. The formula for sides of length "a" and "b" and angle β is ...
... A = (1/2)ab·sin(β)
First note down the relevant variables from the question.
Ua (Initial velocity a) = 320ft/s
Ub (initial velocity b) = 240ft/s
Aay (acceleration of a in the vertical axis) = Aby = -32.17ft/s/s
We want to know when they will be at the same height so should use the formula for displacement:
s = ut + 1/2 * at^2
We want to find when both firework a and firework b will be at the same height. Therefore mathematically when: say = sby (the vertical displacements of firework A and B are equal). We also know that firework B was launched 0.25s before firework A so we should either add 0.25s to the time variable for the displacement formula for firework B or subtract 0.25s for firework A.
SO:
Say = Sby
320t + 1/2*-32.17t^2 = 240(t+0.25) + 1/2 * -32.17(t+0.25)^2
320t - 16.085t^2 = 240t + 60 - 16.085(t+0.25)^2
320t - 16.085t^2 = 240t + 60 - 16.085(t^2 + 0.5t + 6.25)
320t - 16.085t^2 = 240t + 60 -16.085t^2 - 8.0425t - 100.53
320t - 240t - 8.0425t - 16.085t^2 + 16.085t^2 = 60 - 100.53
71.958t = -40.53
t = -0.56s (negative because we set t before Firework A was launched)
Now we know both fireworks explode 0.56 seconds AFTER fireworks B launches (because we added 0.25 seconds to the t variable in the equation above for the vertical displacement of Firework B).
You could continue on to find the displacement they both explode at and verify the answer by ensuring that it is equal (because the question stated they should explode at the same height by substituting the value we found for t of 0.56s into the vertical displacement formula for firework A and t+0.25s=0.81s into the same formula for Firework B
Verification:
Say = ut + 1/2at^2
Say = 320*0.56 + 1/2*-32.17*0.56^2
Say = 179.2 + -5.04
Say = 174.16ft
Sby = ut + 1/2at^2
Sby = 240*0.81 + 1/2*-32.17*0.81^2
Sby = 194.4 - 10.5
Sby = 183.9ft
While Say is close to Sby I would have expected them to be almost perfectly equal… can you please check if this matches the answer in your textbook? There could be wires due to rounding. I also usually work in SI units which use the metric system and not the imperial system although that shouldn’t make a difference. The working out and thought process is correct though and this is why trying to verify the answer is an important step to make sure it works out.
Answer: 0.56s (I think)
Answer:
If I'm correct, it's going to be x > -3
