TheAnointedOne

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From 3:00pm to 4:30pm Vehicle #1 travels a total of 25 mph x 1.5 hours = 37.5 miles. Vehicle #2 catches up at the rate of 30 mph - 25 mph = 5 mph, so it would take 37.5 miles / 5 mph = 7.5 hours to catch up.

Yes you are correct.
 

TheAnointedOne

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extremely easy
4x^3 - 6x^2 + 6x - 5

You just gonna ignore my question, breh? :dahell:

Pythagorean-Theorem.jpg


I know that c = 16 and a = b (since it’s a square), thus
c^2 = a^2 + a^2
Or
c^2 = 2*a^2
thus
16^2=2*a^2
256=2*a^2
128=a^2

'c' goes from the center of the semicircle to the outer rim. It's the radius. So 'c' is 8 not 16.

c^2 = a^2 + a^2
c^2 = 2*a^2
8^2 = 2*a^2
64 = 2*a^2
32 = a^2
4*sqrt(2) = a

Here's some diagrams someone on twitter posted

ECt-3YMWwAAs045


ECuHgOwWkAAFIQX
 
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TheAnointedOne

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Category: Physics -- Mechanics
Difficulty: Easy

You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; the ball is then in free fall. (We ignore air resistance.) On its way back down, it just misses the railing.

At what time after being released has the ball fallen 5.00 m below the roof railing?
 

DrBanneker

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Figthing borg at Wolf 359
Category: Physics -- Mechanics
Difficulty: Easy

You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; the ball is then in free fall. (We ignore air resistance.) On its way back down, it just misses the railing.

At what time after being released has the ball fallen 5.00 m below the roof railing?

In the absence of air resistance, the ball is again going 15 m/s when it reaches the rail due to conservation of potential + kinetic energy. The distance traveled by the ball is then

y = -vt - (1/2)gt^2 with g = 9.8 m/s^2

so

-5 = -15t - 4.9t^2

so we have to solve the quadratic 4.9t^2+15t-5=0
a=4.9, b=15, c=-5

t=[-b+sqrt(b^2-4ac)]/2a

t=[-15+sqrt(225-4*4.9*-5)]/9.8

t=0.3s
 

TheAnointedOne

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In the absence of air resistance, the ball is again going 15 m/s when it reaches the rail due to conservation of potential + kinetic energy. The distance traveled by the ball is then

y = -vt - (1/2)gt^2 with g = 9.8 m/s^2

so

-5 = -15t - 4.9t^2

so we have to solve the quadratic 4.9t^2+15t-5=0
a=4.9, b=15, c=-5

t=[-b+sqrt(b^2-4ac)]/2a

t=[-15+sqrt(225-4*4.9*-5)]/9.8

t=0.3s

Very close but wrong.
 

King

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Math is cool as shyt brehs.

I went up to analytical geometry and calculus in college through sheer force of will.

But I never even learned my times tables in the second grade and pretty much forgot all of math.

This past week I said fukk it, went to the library, and picked up math as a hobby.

Started with the childrens books and could barely get through em. Now I’m in pre-algebra relearning shyt that was force fed to me in school.

It feels good, I was doing long division in my sleep the other night :wow:

I swear imma get back up to multivariable calculus the right way.

Then imma work on some computer science, might even pick up statistics along the way :wow:

Learning is beautiful brehs :mjcry:
 
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