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lesya692 [45]
3 years ago
14

When you have a triangle and 2 of 3 angles are given the degrees and the last one is unknown, you subtract the two given degrees

from each other to get your unknown, right? Just checking :D Tell me if I'm wrong from my notes and file attached please

Mathematics
2 answers:
storchak [24]3 years ago
4 0

No, that's not right.  Sadly, the answer you entered on the
attached drawing is incorrect.  It's slightly more complicated
than that ... only slightly.

First, think about this for a second:  What if the two GIVEN angles
on the drawing had the same number of degrees ?  Then by the
method you've been using, you would subtract them from each
other, and that would give you zero.  So you would say that the
last angle is zero degrees ?  Can you see that this doesn't really
work ?

Here's how it's really done:

It all rests on a rule about triangles.  This is ALWAYS true, and
you should memorize it:

           When you add up the degrees of all three angles
           inside a triangle, the sum is ALWAYS 180 degrees.

So now, when you're given two of the angles, you know that
the unknown one must be exactly enough to bring the sum of
ALL of them up to 180 degrees.

Work it like this:

-- Take the two given angles.
-- ADD them.
-- Subtract their SUM from 180.
   Now you have the third angle.

In the drawing you attached:

-- The given angles are  39  and  102 .
-- Add them:  39 + 102 = 141
-- Subtract the sum from 180:    180 - 141 = 39 .
   The unknown angle is 39 degrees.

But that's the same as one of the given angles ! ?   :-(    ?    :-(

That's OK.  It's perfectly fine for two of the angles, or sometimes
even all three, to be the same size.  They just have to all add up
to 180 degrees, and everything is fine.

Finger [1]3 years ago
4 0
No add then subtract that ans from 180 degrees
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The positions when the particle reverses direction are:

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We are given the following function:

s(t)=2t^{3}-21t^{2}+60t+3

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v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60

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Now, substituting the times to calculate the accelerations, we have:

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Now, substitutitng the times to calculate the positions, we have:

s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft

Have a nice day!

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