Answer:
The value of the ve = 9m/sec
Step-by-step explanation:
From the given formula, it can be conclude that this is evenly accelerated movement.
First I will rewrite given formula
vi = √ ve∧2 - 2ad First we will square on both sides and get
vi∧2 = ve∧2 - 2ad Now we will add monom (+2ad) to both sides and get
vi∧2 + 2ad = ve∧2 -2ad + 2ad =>
ve∧2 = vi∧2 +2ad Now we will rooted both sides and get
ve = √vi∧2 + 2ad
Now we will replace given data vi=7m/sec, a=8m/s∧2 and d=2m in the last formula
ve= √7∧2 + 2*8*2 = √49+32 = √81 = 9
ve= 9m/sec
Good luck!!!
Answer:
The graph in the attached figure
Step-by-step explanation:
we have
-----> equation A
-----> equation B
we know that
The solution of the system of equations is the intersection point both graphs
In this problem
The intersection point is 
therefore
The solution is the point
using a graphing tool
see the attached figure
6x
42/6 = 7 (valid), 18/6 = 3 (valid), 12/6 = 2 (valid)
7, 8, 9, 10, and 11 are all not divisible by 12, and 12 is not divisible by 18, so 6 is the highest number.
Also, each number has at least one x, so you can take one out.
Remaining=
6x (7 + 3x^2 + 2x^3)
9514 1404 393
Answer:
15.71 units
Step-by-step explanation:
The arc length is given by the formula ...
s = rθ
where r represents the radius, and θ represents the central angle in radians.
The angle of 60° is π/3 radians, so the arc length is ...
s = 15(π/3) = 5π ≈ 15.71 . . . units
LN ≈ 15.71 units
__
<em>Note</em>: if you use 3.14 for π, you will get 15.70.
Answer:
Step-by-step explanation:
Let many universities and colleges have conducted supplemental instruction(SI) programs. In that a student facilitator he meets the students group regularly who are enrolled in the course to promote discussion of course material and enhance subject mastery.
Here the students in a large statistics group are classified into two groups:
1). Control group: This group will not participate in SI and
2). Treatment group: This group will participate in SI.
a)Suppose they are samples from an existing population, Then it would be the population of students who are taking the course in question and who had supplemental instruction. And this would be same as the sample. Here we can guess that this is a conceptual population - The students who might take the class and get SI.
b)Some students might be more motivated, and they might spend the extra time in the SI sessions and do better. Here they have done better anyway because of their motivation. There is other possibility that some students have weak background and know it and take the exam, But still do not do as well as the others. Here we cannot separate out the effect of the SI from a lot of possibilities if you allow students to choose.
The random assignment guarantees ‘Unbiased’ results - good students and bad are just as likely to get the SI or control.
c)There wouldn't be any basis for comparison otherwise.
