You put .875 over the number 1 to make it a fraction so 
1 answer:
The LCM of the given numbers have been determined.
<h3>What is LCM ?</h3>LCM stands for Least Common Multiple .
For two numbers , LCM is the number that is the smallest number of which they both are a factor.
It is asked to determine
LCM of 6 and 9
6 = { 6 , 12 , 18 , 24 , 30 , 36 , 42 .....}
9 = { 9 , 18 ,27 .....}
The LCM of 6 , 9 is 18
LCM of 8 and 12
8 = { 8,16,24,32....}
12 = {12 , 24 ,36....}
The LCM of 8 and 12 is 24
LCM of 4 , 6 and 8
4 = {4 , 8 , 12 , 16 , 20,24,28 ....}
6 = { 6 ,12 ,18,24......}
8 = {8 , 16,24 .....}
The LCM of 4,6 and 8 is 24
The LCM of 6 and 15 is 30
The LCM of 20 and 30 is 60
The LCM of 25,30 and 75 is 150
The image of the solution is attached .
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Answer:
1)
Minimum is a 4th Degree
2)
Positive; Even
3)
Step-by-step explanation:
Part 1)
The minimum degree of our function will be 4.
Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it <em>crosses through</em> the x-axis at these two points, these two factors must have an odd multiplicity.
So, it can be anything 1, 3, 5, 7, etc.
However, we will choose the lowest one, 1.
Next, we know that the graph <em>bounces off</em> at 3.
So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.
We choose the lowest one, 2.
Therefore, the minimum degree of our function will be 1+1+2 or 4.
Part 2)
The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.
Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.
This applies to all polynomials with even degrees.
Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.
Part 3)
This is similar to Part 1.
We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is:
We know that it<em> passes through</em> . So, these two factors must have an odd multiplicity.
However, since the graph <em>bounces off</em> , this factor must have an even multiplicity.
And we're done!