Posted to your other question as well:
So we have:
A)
B)
C)
D)
We need to try x = –6 in each equation until we find a solution that makes both sides of the equation equal.
For A, this gives an undefined result. Logarithms are only defined for bases that are positive real numbers not equal to 1.
For B, this again gives an undefined result. The logarithm of a number is only defined when that number is a positive real number. It would be –21, which is undefined.
For C, we have a defined operation, but
.
Finally, for D, we have
This is true.
Answer:
4n-1
Step-by-step explanation:
The difference in the numbers is 4 so the coefficient must be 4 for the nth term meaning
4n
11 is the third term however we need to find the first so..
11=3rd term
-4
7=2nd term
-4
3=1st term
now we need to minus 4 from 3 which is -1 (3-4)
meaning the answer put together will be 4n-1
Answer:
like you wanna message on brainly? or...something else
This problem is a real bear. Whoever wrote it has a sense of humor, and expects
a lot from you ... probability, permutations, combinations, and geometry.
<u>First, let's talk about probability:</u>
The probability of something happening is (the number of outcomes that meet your description) divided by (the total number of all possible outcomes).
<u>#62:</u>
There are 4 points on the drawing. How many different ways could you pick
two of them ?
The first one you pick could be any one of 4 points.
For each of those, the other point could be any one of the remaining 3.
So the total number of ways to pick 2 points out of 4 is (4 x 3) = 12 ways.
But wait ! Whether you pick 'A' and then 'C', or pick 'C' and then 'A', you still wind up with the same two points. So, although there are 12 ways to pick them, there are only 6 different distinct pairs of points.
OK. How many of those pairs are collinear ? ANY two points lie on the same line, because a line can always be drawn between any two points. So out of the 6 different possible pairs of points, ALL 6 pairs are collinear. The probability of picking a pair that are collinear is 100% .
<u>#63:</u>
There are 4 points on the drawing. How many different ways could you pick
three of them ?
The first one you pick could be any one of 4 points. For each of those ...
The second point could be any one of the remaining 3. For each of those ...
The third point could be either one of the remaining 2.
So the total number of ways to pick 3 points out of 4 is (4 x 3 x 2) = 24 ways.
But
wait ! Whether you pick ABC, ACB, BAC, BCA, CAB, or CBA, you still wind up with the same three points. So, although there are 24
ways to pick them, there are only 4 different distinct sets of three points.
OK. How many sets of 3 points in this drawing are collinear ?
There is only one ! ONLY A, B, and C are collinear.
Probability = 1 out of 4 = 1/4 = 25 percent .
<u>#64:</u>
In the last problem, we saw that there are 4 distinct sets of three points.
How many of them are coplanar ?
They ALL are. A plane can be drawn through ANY three points.
So whichever three points you pick, they are coplanar.
The probability is 4 out of 4 = 100 percent.
Thank you for your 5 points. I shall cherish them.