John has $5 in his bank account when he gets a job. He begins making $109 a day. A student found that the equation that represen
1 answer:
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Answer:
B: 8 7/8 Lbs
Step-by-step explanation:
answer =
step 1 find a common denominator
common denominator - 8
* apply *
now we have
step 2 make each fraction a improper fraction
We can do this my multiplying the big number by the denominator and adding that to the numerator.
now we have
Now we can subtract
remember when subtracting fractions you only subtract the numerators. ( so we keep the denominator as 8 )
so 121 - 28 - 22 = 71
so the answer is
finally we want to change the answer back into a mixed fraction
To do this we divided 71 and 8
8 can go into 71 8 times
8 * 8 = 64
71 - 64 = 7
We're left with
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!