100
explanation: those two angles are supplementary which add up to be 180.
Answer: 5x^(-1) or 5/x.
Step-by-step explanation: The equation given is (5*x^2)/x^3 and we are supposed to solve it for x = 2. First. we simplify the equation. Since we have x^2 in the numerator and x^3 in the denominator, using the equation a^x/a^y = a^(x - y) we find that we get 5*x^(-1) we can leave it like this or simplify it to 5/x, since negative exponents mean fractions.
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:
359 + r = 212
Step-by-step explanation:
"359 and r more" suggests that you are adding the number 359 with the variable r. r + 359
"equals 212" is "= 212"
Put the two together:
359 + r = 212 is your answer.
~
Answer:
The answer is below
Step-by-step explanation:
The bottom of a river makes a V-shape that can be modeled with the absolute value function, d(h) = ⅕ ⎜h − 240⎟ − 48, where d is the depth of the river bottom (in feet) and h is the horizontal distance to the left-hand shore (in feet). A ship risks running aground if the bottom of its keel (its lowest point under the water) reaches down to the river bottom. Suppose you are the harbormaster and you want to place buoys where the river bottom is 20 feet below the surface. Complete the absolute value equation to find the horizontal distance from the left shore at which the buoys should be placed
Answer:
To solve the problem, the depth of the water would be equated to the position of the river bottom.
