Answer:
(8, 11 )
Step-by-step explanation:
Given the 2 equations
2x + 2y = 38 → (1)
y = x + 3 → (2)
Substitute y = x + 3 into (1)
2x + 2(x + 3) = 38 ← distribute and simplify left side
2x + 2x + 6 = 38
4x + 6 = 38 ( subtract 6 from both sides )
4x = 32 ( divide both sides by 4 )
x = 8
Substitute x = 8 into (2) for corresponding value of y
y = 8 + 3 = 11
Solution is (8, 11 )
The rate of change in this pattern is 2:1 comparing the change in 'y' to the change in 'x'.. this means that if you go form 33 to 51 ( an increase of 18), then the the increase from 16 to ? will be 9. Therefore, ? = 16 + 9 or 25
? = 25
There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:
Consider an athlete running a 40-m dash. The position of the athlete is given by
where d is the position in meters and t is the time elapsed, measured in seconds.
Compute the average velocity of the runner over the intervals:
(a) [1.95, 2.05]
(b) [1.995, 2.005]
(c) [1.9995, 2.0005]
(d) [2, 2.00001]
Answer
(a) 6.00041667m/s
(b) 6.00000417 m/s
(c) 6.00000004 m/s
(d) 6.00001 m/s
The instantaneous velocity of the athlete at t=2s is 6m/s
Step by step Explanation:
In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:
so let's take the first interval:
(a) [1.95, 2.05]
we get that:
so:
(b) [1.995, 2.005]
we get that:
so:
(c) [1.9995, 2.0005]
we get that:
so:
(d) [2, 2.00001]
we get that:
so:
Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s
The number would be greater. if you think about normal multiplication, when you multiply a number by 1 the number stays the same, but when you multiple the number by less then one, i.e. a fraction you get a fraction of what you had. while a number greater then one, i.e. 2, the number increases. so 1.23 is greater then one so it would increase
The correct answer is B. 10,24,26