Here's our equation.

We want to find out when it returns to ground level (h = 0)
To find this out, we can plug in 0 and solve for t.


So the ball will return to the ground at the positive value of

seconds.
What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

As we can see that the sides of the rectangle have been doubled
6.2 ft changed to 12.4 ft
Now when the sides have been doubled
the Perimeter will also be doubled
So the perimeter of new rectangle should be the double the perimeter of old rectangle
So perimeter of new rectangle = 2 (16 ) = 32 feet
Option C is correct
59.9%
Sorry I don’t have time to explain in the middle of an exam
Answer:
30
Step-by-step explanation:
If they are parallel then the 2 labelled angles sum to 180.
2x+50+5x-80=180
7x-30=180
7x=210
x=30
Answer:
15. ∠ABE = 50°
16. ∠EBD = 10n - 19
Step-by-step explanation:
If
bisects ∠ABD then we have ∠ABE = ∠DBE
So, we will have: (6x + 2)° = (8x - 14)°
⇒ 6x + 2 = 8x - 14
⇒ 8x - 6x = 14 + 2
⇒ 2x =16
⇒ x = 8
∴ ∠ABE = 6(8) + 2
= 48 + 2
= 50°
∴∠ABE = 50°
16. If
bisects ∠ABD then ∠ABE + ∠EBD = ∠ABD
⇒(12n - 8) + ∠EBD = (22n - 11)
⇒ ∠EBD = 22n - 11 - 12n + 8
⇒ ∠EBD = 10n - 3
Hence, the answer.