Answer:
<em>A) (-5,7)</em>
Step-by-step explanation:
<u>Functions and Relations</u>
A set of values A can have a relation with another set B as long as at least one element of A has at least one image in B. Functions are special relations where each element of A (the domain of the function) has one and only one image on B (the range of the function).
By looking at the options, we can see that x=9, x=-8, and x=-1 already have defined values in Y, so if we define another value for any of them the relation will stop being a function. The only possible choice to preserve the function is the option
If there are 8 marbles in a group , and 5 out of every 8 is red, than in six groups there would be 5, 10, 15, 20, 25, 30 red marbles.
This lists the group of 6 sets.
30 red marbles in 6 sets.
Answer:
1:4
Step-by-step explanation:
Let's find the ratio of one side of Figure A to one side of Figure B. Note that the have to be the same side on each triangle (Ex: The short side and the short side or the medium length side and the medium length side or the long side and the long side)...
14:56
We can simplify this into...
1:4
Answer:
x = 2
Step-by-step explanation:
Taking antilogs, you have ...
2³ × 8 = (4x)²
64 = 16x²
x = √(64/16) = √4
x = 2 . . . . . . . . (the negative square root is not a solution)
___
You can also work more directly with the logs, if you like.
3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2
3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents
6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify
2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)
ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2
2 = x . . . . . . . . . . . . . . . . . . . take the antilogs
I believe the answer your looking for is B.. good luck my friend!
