Answer:
its B.
Step-by-step explanation: just take half off the numbers, and then you have your answer. simple.
Answer:
Graph (C)
Step-by-step explanation:
To find whether the graph\table represents a relationship or a function we have to analyze the input-output values given.
Graph A.
In this graph for every input value (x-value) there are two output values (y-values).
For x = -2, y = -2, 2
So the graph doesn't represent a function.
Graph B.
For every x value there are two y-values
For x = -5, y = -3, 3
So the graph doesn't represent a function.
Graph C.
For every input value there is a different y-value.
Therefore, graph represents a function.
Graph D.
In this graph for x = 3, y = 1, 2, 3, 4
For one value of x, there are four values of y.
Therefore, graph doesn't show the relationship.
I think I may be wrong check
5(x^2n-1)×(2x^3n-1) ^2
=20n3 x8 -20n2 x5 +20n x3 +20n x3 +5n x2-5
Answer:
x>6
Step-by-step explanation:
2x > 16-4
x > 12/2
x > 6
Sorry I am a little late...
a = -2
b = -9
Here is how to solve the problem.
First thing I did was multiply the first equation by -2 so that we can eliminate the the b. After you multiply it by -2, your new equation is -16a + 8b = -40.
You leave the second equation alone and all you do is combine like terms. So -16a+5a is -11. And you eliminate the b. Then you're going to do -40+62 which is 22. So it's -11a=22 and then you have to solve for a. What I did was I multiplied the whole thing by minus to turn the a positive. So then it's 11a=-22. Pretty easy, the final step is to simplify. -22/11 is -2. ;D
So there you have your first answer.
a = -2
Now we're going to use the first answer to help us find b.
For the second equation, all you're going to do is plug in that a.
5 (-2)-8b=62
-10 - 8b = 62
Now we move the -10 to the other side...
-8b = 62 + 10
-8b = 72
Multiply the whole thing by negative once again to turn the b positive.
Now we have 8b = -72
The final step is to simplify. -72/11 = -9
b = -9
Hope this makes sense! Also I had the same question on my test and I got it right. :)