For this we will use kinematic equations.
We need the time of flight to compute the range.
With the given information, the most useful formula is:
Δy=v0yt−g2t2
Because our equation is taking only vertical motion into account,
v0y=30sin(45)
Because we are looking for the range of the object,
Δy=0
This leaves us with only one unknown: t
0=30sin(45)t−g2t2
g2t2=30sin(45)t
g2t=30sin(45)
t=60gsin(45)
For horizontal distance, the formula we need is:
Δx=v0xt
Because we are taking into account the horizontal range,
v0x=30cos(45)
In substituting our derived values into the equation we can see that:
Δx=30cos(45)∗60gsin(45)
Range=30cos(45)∗60gsin(45)
Answer:
<u>x = 7</u>
Explanation:
Because 25 is a perfect square of 5, we can turn
= 
into
= 
Since the bases are now both equal, we can completely ignore them, as we are only trying to find x. This leaves us with:
3x - 5 = 2x + 2
All we have to do now is solve for x:
x - 5 = 2 <em>Subtract 2x from both sides.</em>
<u>x = 7</u> <em>Add 5 to both sides.</em>
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<em>Hope this helps! :)</em>
Answer:
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You know the answer is correct when the answer has a good star review such as 5 stars or has a good amount of likes. Everyone has a title by their name, the higher status it is, the more likely their answer would be right.