<em>Answer:</em>
<em>80 books = $400</em>
<em>Step-by-step explanation:</em>
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Answer:
Step-by-step explanation:
Each signal unit cubes are 1 cm to each side
the dimensions of the large cube is 4 by 6 by 3 width height and depth
a) volume of the cube Vol = area of the base times height
area of the base = the width times the depth
Volume = Base Depth
= Width x Depth x Height
= 4 x 3 x 6
= 12 x 6
Vol = 72 cm ³
b ) the volume of the smaller cuboid has different dimensions
the smaller cube is 2 by 2 by 1 width height and depth
Volume = Base Depth
= Width x Depth x Heigh
= 2 x 2 x 1
Vol smaller cuboid = 4
How many smaller cubes can be made?
The obvious answer might be to straight up divide the large cube volume by the smaler cuboid volume. That might work IF all of the cuboid dimensions divide evenly into the cube dimensions.
Number of cuboid = Volume of the cube / Volume of the cuboid
= 72 / 4
= 18
Check this answer by dividing the dimension of the cube by the cuboid
4 by 6 by 3
2 by 2 by 1
(4/2) by (6/2) by (3/1)
2 by 3 by 3 = 2 x 3 x 3 = 18
x-3x^3
Step-by-step explanation:
just pretend they are in parentheses and open them
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)
