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arieannaensley0616
02.10.2019 •
Mathematics
An object which is vertically projected or falls from the sky has a velocity of 'v' and is given by: v = -32t + vo, where 'v' is the velocity in feet/sec, 't' is the time in seconds, and 'vo' is the initial velocity in feet per second. if the initial velocity is 195 feet/sec, and the velocity v is in the range of 3 < v < 99, find the time interval t.
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Ответ:
3 = -32t + 195
-32t = 3 - 195
t = -192 / -32
t = 6
Greatest Range = 99 = -32t + 195
-32t = 99 - 195
t = -96 / -32
t = 3
In short, Your Time Interval would be: 3 < t < 6
Hope this helps!
Ответ:
14.
15.
Step-by-step explanation:
Solving (14):
First, we solve for x.
Since sides
, then
This is so because the theorem of ASA (Angle-Side-Angle)
So, we have:
Collect Like Terms
Divide both sides by 3
Solving for m<B, m<C and m<D.
We simply substitute 11 for x in their respective expressions.
Solving (15):
First, we solve for x.
Since
, then
This is so because the theorem of SAS (Side-Angle-Side)
So, we have:
Collect Like Terms
Divide both sides by 2
Solving for WX, WY and XY
We simply substitute 4 for x in their respective expressions.