Answer:
9,32 . 7,16 . 3,8 . 5,8 . 3,4 . 1,2
in that order buddy make it exactly like this :)
<h3>Answer:</h3>
The point (4, -3) is located in Quadrant IV (4). The point is four spots to the right, and 3 spots downwards, putting it in Quadrant 4.
i think the answer is A
Step-by-step explanation:
i just think it's a
Given that a flyer begins from earth and thrown up with a velocity of 30 ft/sec vertically.
u = initial velocity = 30 : a = -g = 32 ft/sec^2
s(0) =initial height =4 ft.
We have the equation
where u = initial velocity : v= final velocity : s = distance travelled and a = acceleration. Here final velocity is found out as follows
Substitute to get
Since v cannot be negative, the flyer’s center of gravity cannot ever reach 20 feet.
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To reach 25 ft, put s = 25
and v must be atleast 0
Then we have
Hence if thrown with initial velocity of 40 ft /sec, it will reach a height of 25 ft.
Answer:
4m²n²
Step-by-step explanation:
you have to basically mutiply everything by itself
4m²n²