Answer:
In words the answer is between t=0 and t=2.
In interval notation the answer is (0,2)
In inequality notation the answer is 0<t<2
Big note: You should make sure the function I use what you meant.
Step-by-step explanation:
I hope the function is h(t)=-16t^2+32t because that is how I'm going to interpret it.
So if we can find when the ball is on the ground or has hit the ground (this is when h=0) then we can find when it is in the air which is between those 2 numbers.
0=-16t^2+32t
0=-16t(t-2)
So at t=0 and t=2
So the ball is in the air between t=0 and t=2
Interval notation (0,2)
Inequality notation 0<t<2
Answer:
x = 27, y = 74
Step-by-step explanation:
Hi again :)
Since the angles of 5x + 4 and the one adjacent to x = 14 are corresponding angles, they are equal. That means that (5x + 4) + (x + 14) = 180 degrees.
5x + 4 + x + 14 = 180
6x + 18 = 180
6x = 162
x = 27
Also, the angle (5x + 4) and (2y - 9) are vertical angles, they are equal in value. We know the value of x now, so we'll substitute that in and solve for y.
5(27) + 4 = 2y - 9
135 + 4 = 2y - 9
139 = 2y - 9
148 = 2y
y = 74
As always, lmk if you have questions.
Answer:
the correct choice is marked
Step-by-step explanation:
Let x represent the smaller number. Then the larger number is 8x, and the difference is ...
8x -x = 280
7x = 280 . . . . . simplify
x = 40 . . . . . . divide by 7
The larger number is 8x = 8(40) = 320.
_____
<em>Additional comment</em>
Effectively, we have solved for the multiplier (40) that gives the ratio with 320 on top and a difference between top and bottom of 280:
Answer:
g(6) = 71
f(11) = 62
Step-by-step explanation:
Let's solve g(6) first
Plug 6 into x
g(6) = 2(6)^2 - 1
g(6) = 2(36) - 1
g(6) = 72 - 1
g(6) = 71
Now let's solve f(11)
f(11) = 5(11) + 7
f(11) = 55 + 7
f(11) = 62
<em>Thus, out answers are 71 and 62 respectively</em>
