Answer:
c
Step-by-step explanation:
used socratic
Its just a matter of subbing in ur answer choices to see which ones are correct...but remember, for it to be a solution, it has to satisfy BOTH equations.
(-1/4,-4)
y = 8x - 2
-4 = 8(-1/4) - 2
-4 = - 8/4 - 2
-4 = -2-2
-4 = -4 (correct)
(-1/4,-4)
y = -4x - 5
-4 = -4(-1/4) - 5
-4 = 1 - 4
-4 = -4 (correct)
Therefore, ur solution is (-1/4,-4)
Given the following group of numbers, 8, 2, 9,4, 2, 7, 8, 0, 4, 1, which of the following
Lena [83]
Answer:
III. The mode is 2
Step-by-step explanation:
Mean: (8+2+9+4+2+7+8+0+4+1)/10 = 45/10 = 4.5
Median: (2+7)/2 = 9/2 = 4.5
Mode: There is no mode since 2 and 8 both appear twice (in other words, since they have the same frequency, it's bimodal)
The closest possible choice is III. The mode is 2, but keep in mind that since the data set is bimodal, there is no single mode.
Answer(there are assumptions for this answer that you need to confirm and look at):
Assumptions:
and 
Answer if the operation is multiplication:
If you meant a closed dot which is the symbol for multiplication.


Answer if the operation is composition:
If you meant an open dot which is the symbol for composition.


Note: I don't know if you actually meant
or if
. Please let me know one way or the other.
Step-by-step explanation:
If we assume the functions are:


since multiplication is commutative.






We are asked to find
and
.
The order doesn't matter in multiplication.








Now you might have meant composition which symbolized with an open circle, not a closed one.


since 





since 


