Answer:
t = 2.52 seconds
Step-by-step explanation:
h=139-15t-16t^2
We want to know when the ball hits the ground
That would be when h=0
0 = 139-15t-16t^2
We can use the quadratic formula to find t
t = -b ± sqrt(b^2-4ac)
----------------------
2a
where a = -16 b = -15 and c = 139
t = -(-15) ± sqrt((-15)^2-4(-16)139)
----------------------
2(-16)
t = (15) ± sqrt(225+8896)
----------------------
-32
t = (15) ± sqrt(9121)
----------------------
-32
t = 15+ sqrt(9121) t = 15- sqrt(9121)
-------------------- or -------------------
-32 -32
-3.453247707 or 2.515747707
Since time cannot be negative
2.515747707
Round to the nearest hundredth
t = 2.52 seconds
Y^2 -11y = 0
We can factor it:
y * (y -11) = 0
So y = 11, 0
An infinite amount...since there ARE and infinite amount of number in the universe.
1111
2222
3333
4444
5555
6666
7777
8888
9999
1010
1234
4321
5432
2345
5678
8765
0987
7890
4509
9054
Etc
9514 1404 393
Answer:
40°
Step-by-step explanation:
Triangles QRS and QTS are congruent (HL), so the marked angles are also congruent:
3x +2 = 4x -4
6 = x
Then the total angle measure of angle RST is ...
(3x +2) +(4x -4) = 7x -2 = 7(6) -2 = 40 . . . degrees
m∠RST = 40°
If Susan had 2$ and got to 12$ we need to find what multiplied by 2 = 12. So, 12/2=6
If 6x2=12 then 11x6=66
66$
