Answer:
The inverse for log₂(x) + 2 is - log₂x + 2.
Step-by-step explanation:
Given that
f(x) = log₂(x) + 2
Now to find the inverse of any function we put we replace x by 1/x.
f(x) = log₂(x) + 2
f(1/x) =g(x)= log₂(1/x) + 2
As we know that
log₂(a/b) = log₂a - log₂b
g(x) = log₂1 - log₂x + 2
We know that log₂1 = 0
g(x) = 0 - log₂x + 2
g(x) = - log₂x + 2
So the inverse for log₂(x) + 2 is - log₂x + 2.
16/17 is the answer. you just add the numerators since they have common denominators!
Answer:
K = 43
Step-by-step explanation:
We'll begin by determining the gradient of the equation 5y + 4x = 8. This can be obtained as follow:
5y + 4x = 8
Rearrange
5y = 8 – 4x
5y = –4x + 8
Comparing 5y = –4x + 8 with y = mx + c, the gradient m is –4
Next, we shall determine the gradient of the line perpendicular to the line with equation 5y = 8 – 4x.
This can be obtained as follow:
For perpendicular lines, their gradient is given by:
m1 × m2 = – 1
With the above formula, we can obtain the gradient of the line as follow:
m1 × m2 = – 1
m1 = –4
–4 × m2 = – 1
Divide both side by –4
m2 = –1/–4
m2 = 1/4
Finally, we shall determine the value of k as follow:
Coordinate => (k, 4) and (3, –6)
x1 coordinate = k
y1 coordinate = 4
x2 coordinate = 3
y2 coordinate = –6
Gradient (m) = 1/4
m = (y2 – y1) / (x2 – x1)
1/4 = (–6 – 4) / (3 – K)
1/4 = –10 /(3 – K)
Cross multiply
3 – K = 4 × –10
3 – K = –40
Collect like terms
– K = – 40 –3
–k = –43
Divide both side by – 1
K = –43/–1
k = 43
A circles DIAMETER is twice as long as its radius
Hope this helps you :)
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
