Answer:
the two roots are x = 1 and x = 4
Step-by-step explanation:
Data provided in the question:
(x³ − 64) (x⁵ − 1) = 0.
Now,
for the above relation to be true the following condition must be followed:
Either (x³ − 64) = 0 ............(1)
or
(x⁵ − 1) = 0 ..........(2)
Therefore,
considering the first equation, we have
(x³ − 64) = 0
adding 64 both sides, we get
x³ − 64 + 64 = 0 + 64
or
x³ = 64
taking the cube root both the sides, we have
∛x³ = ∛64
or
x = ∛(4 × 4 × 4)
or
x = 4
similarly considering the equation (2) , we have
(x⁵ − 1) = 0
adding the number 1 both the sides, we get
x⁵ − 1 + 1 = 0 + 1
or
x⁵ = 1
taking the fifth root both the sides, we get
![\sqrt[5]{x^5}=\sqrt[5]{1}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E5%7D%3D%5Csqrt%5B5%5D%7B1%7D)
also,
1 can be written as 1⁵
therefore,
![\sqrt[5]{x^5}=\sqrt[5]{1^5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E5%7D%3D%5Csqrt%5B5%5D%7B1%5E5%7D)
or
x = 1
Hence,
the two roots are x = 1 and x = 4
F(x) = -5/6x - 3
You can get this by identifying the y-intercept and then using the formula for slope given any two points on the graph.
Hello!
You have to do 7 * 2/5
Which is the same as 7/1 * 2/5
Which gives us the answer 2 4/5 bottles of glue.
Hope this helps!
Answer:
yes
Step-by-step explanation:
Answer:
see attached
Step-by-step explanation:
The grid show the non-possibilities in red, with each number corresponding to the statement that eliminates that choice. The green square (with black text) shows the one combination that is specified already (by statement 4). The lighter green numbers show possible alternatives: first period may be Schiller or English, and room 113 will be the other one. Similarly, Art may be 3rd period or Thomlinson.
__
These choices (light green 5, light green 6) give rise to four possibilities. Working through them, you run into inconsistencies if you choose Schiller for first period. (Art must be, but can't be, in room 112.) That leaves two possibilities.
Again, you run into inconsistencies if you choose Thomlinson as the Art teacher. (The class in 112 is 2 periods after Xavier's class, not 1.)
Hence, the only viable pair of remaining choices is Schiller in room 113 and art in 3rd period.
The final schedule is shown in the attachment.
_____
<em>Additional comment</em>
When I'm working these on paper, I use an X to mark any impossible combinations, and a circle (O) to mark a known combination. In any given 4×4 square of the grid, the remaining cells of the row and column containing a O must be Xs. Consistency must be maintained between rows and columns. This often means filling a circle in one place may result in a circle being filled in another place. Of course, once 3 of the squares in a row or column have Xs, the remaining one must be O.
