Answer:
After 3 seconds while going up and 6 seconds while coming down
Step-by-step explanation:
Since projectile is launched from ground level with an initial velocity of v 0 feet per second, initial height =0
Height is given by
where v0 = initial velocity = 144
a) When h =288 ft.
At t= 3 or 6 seconds the projectile would be at height 288 ft.
So while going up after 3 seconds it wouldbe at a height of 288 ft and while coming after 6 seconds.
Answer:
HN
Step-by-step explanation:
I'm going to assume that the room is a rectangle.
The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.
You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.
The dimensions of your room are 25ft (length) by 24ft (width).
Answer:
the two angles are 55° each
Step-by-step explanation:
Let the two equal angles be y each
Therefore,
70° + y + y = 180°
Collect like terms
2y = 180 — 70
2y = 110
Divide both side by the coefficient of y i.e 2
y = 110/2
y = 55°
Therefore, the two angles are 55° each
Answer:
attached images
Step-by-step explanation:
