Answer:
D) 14 seconds
Step-by-step explanation:
First we will plug 500 in for y:
500 = -4.9t² + 120t
We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:
500-500 = -4.9t² + 120t - 500
0 = -4.9t²+120t-500
Our values for a, b and c are:
a = -4.9; b = 120; c = -500
We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.
The quadratic formula is:
Using our values for a, b and c,
The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is
19-5 = 14 seconds.
(3x + 14) (3x + 14) = 9x^2 + 24x + 16
You have to add the problem
The answer 0.1, 0.136, 0.652, 0.87
To find the 20th term in this sequence, we can simply keep on adding the common difference all the way until we get up to the 20th term.
The common difference is the number that we are adding or subtracting to reach the next term in the sequence.
Notice that the difference between 15 and 12 is 3.
In other words, 12 + 3 = 15.
That 3 that we are adding is our common difference.
So we know that our first term is 12.
Now we can continue the sequence.
12 ⇒ <em>1st term</em>
15 ⇒ <em>2nd term</em>
18 ⇒ <em>3rd term</em>
21 ⇒ <em>4th term</em>
24 ⇒ <em>5th term</em>
27 ⇒ <em>6th term</em>
30 ⇒ <em>7th term</em>
33 ⇒ <em>8th term</em>
36 ⇒ <em>9th term</em>
39 ⇒ <em>10th term</em>
42 ⇒ <em>11th term</em>
45 ⇒ <em>12th term</em>
48 ⇒ <em>13th term</em>
51 ⇒ <em>14th term</em>
54 ⇒ <em>15th term</em>
57 ⇒ <em>16th term</em>
60 ⇒ <em>17th term</em>
63 ⇒ <em>18th term</em>
66 ⇒ <em>19th term</em>
<u>69 ⇒ </u><u><em>20th term</em></u>
<u><em></em></u>
This means that the 20th term of this arithemtic sequence is 69.
