It’s basically 100 multiplied by 2 so 200
        
             
        
        
        
Answer:
5
Step-by-step explanation:
thanks for the points
 
        
             
        
        
        
Width = 5, length = 9
Area = length * width
length = (2 * width) - 1
sub equation for length 
45 = (2w - 1) * w
45 = 2w^2 - w
2w^2 - w - 45 = 0
(2w - 9)(w - 5)
2w - 9 = 0
w = 9/2, l = 8 no
w -5 = 0
w = 5, l = 9 yes
        
             
        
        
        
Answer:
the slope would be rise over run
it is shown in the graph it rise 3units and run 2units therefore it's 3/2
 
        
             
        
        
        
Answer:
(a) The maximum height of the ball is 16 feet at 1 seconds.
(b) The ball hits the ground at 2 seconds.
PART (A) :
Given function: h(t) = -16t² + 32t
Comparing to quadratic function: ax² + bx + c
In this function: a = -16, b = 32, c = 0
<u>To find the maximum height</u>, use the vertex formula:

<u>Insert values</u>

Then find h(t) = -16(1)² + 32(1) = 16 feet
Conclusion: The maximum height of the ball is 16 feet at 1 seconds.
PART (B) :
When the ball hits the ground, the height [h(t)] will be 0 ft
-16t² + 32t = 0
-16t(t - 2) = 0
-16t = 0, t - 2 = 0
t = 0, t = 2
The ball hits the ground at 2 seconds. Note: the other 0 seconds is for when the ball was launched at the beginning.