The function involves the initial speed of the object, and the height
from which it was launched, and whether it was launched upward
or downward. As you can see, your post was corrupted and is
unreadable. But when the object hits the ground, its height is zero.
Set the function equal to zero, solve the quadratic equation that you
have, discard the negative solution, and the positive solution is your
answer.
Step-by-step explanation:
are in the solution:
(3,1)
(-1,2)
(6,0)
(0,0)
are not
(-2,-1)
(0,-4)
A..Just plus it if not wrong
Area of square = s^2
Area of Rectangle = lw
l = 2s
w+3 = s
solve for w, w=s-3
Now we have l and w.
Plug both into area of rectangle formula so: (2s)(s-3)
Since both areas are equal set both equations equal to each other:
(2s)(s-3)=s^2
Now simplify
2s^2-6s=s^2
s^2-6s=0, Solve for s.
Factor polynomial. s(s-6)=0 , s can be equal to 0 or 6. HOWEVER, you cannot have a side length of 0 therefore the side length has to be 6.
Now plug in s for the length formula for the rectangle:
l = 2s so... l = 2(6) so length of rectangle = 12.
Now plug in s for the width formula for the rectangle:
w+3=s so... w+3=6 so width of rectangle = 3.
Now the dimensions of the rectangle are 12 by 3. 12 being length and 3 width.
To CHECK:
Find area of rectangle:
A=lw so A=3 times 12 so A=36
Find area of square:
We know the side is equal to 6 so
A=s^2 so 6^2 = 36
The areas are equal that verifies the answer of 12 by 3.
Answer:
(3×3)+2=11
$11
Step-by-step explanation:
