Simple: 3 shelves times 40 games = 120 games, 120-85= 35. So the answer is 35 games.
Answer:
19ft
Step-by-step explanation:
Given the height of a ball above the ground after x seconds given by the quadratic function y = -16x2 + 32x + 3, we can find the maximum height reached by the ball since we are not told what to look for.
The velocity of the ball is zero at maximum height and it is expressed as:
V(x) = dy/dx
V(x) = -32x+32
Since v(x) = 0
0 = -32x+32
32x = 32
x = 32/32
x = 1s
Get the height y
Recall that y = -16x² + 32x + 3.
Substitute x = 1
y = -16(1)²+32(1)+3
y = -16+32+3
y = -16+35
y = 19ft
Hence the maximum height reached by the ball is 19ft
Answer:
The first one
Step-by-step explanation:
g(x) = ax² + bx + c
Point (0,0):
0 = a.0² + b.0 + c
c = 0
Point (2,1):
1 = a.2² + b.2 + c
4a + 2b + c = 1
But c = 0. Then:
4a + 2b = 1
Another point: (- 2, 1):
1 = a.(- 2)² + b.(- 2) + c
4a - 2b = 1
{4a + 2b = 1
{4a - 2b = 1
4a + 2b = 4a - 2b
4a - 4a = - 2b - 2b
- 4b = 0
b = 0
4a + 2b = 1
4a + 2.0 = 1
4a = 1
a = 1/4
The formula is:
g(x) = (1/4)x²
I hope I've helped you.
Answer:
ΔNAS≅ΔSEN by SSA axiom of congruency.
Step-by-step explanation:
Consider ΔNAS and ΔSEN,
NS=SN(Common ie . Both are the same side)
SA=NE( Given in the question that SA≅ NE)
∠SNA=∠NSE( Due to corresponding angle property where SE ║ NA)
Therefore, ΔNAS ≅ΔSEN by SSA axiom of congruency.
∴ NA≅SA by congruent parts of congruent Δ. Hence, proved.
