<h3>Answer: Choice C. </h3><h3>Division[ (4x^3+2x^2+3x+5)^2, x^2+3x+1]</h3>
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Explanation:
It honestly depends on the CAS program, but for GeoGebra for instance, the general format would be Division[P, Q]
Where,
- P = numerator = (4x^3+2x^2+3x+5)^2
- Q = denominator = x^2+3x+1
As another example, let's say we want to divide x^2+5x+6 all over x^3+7 as one big fraction
We would type in Division[x^2+5x+6, x^3+7]
Answer:
Step-by-step explanation:
3/(x-2) - 2/(x+3) = 1....multiply everything by (x-2)(x+3)
3/(x-2) * (x-2)(x+3) - 2/(x+3) * (x-2)(x+3) = 1(x-2)(x+3)
3(x+3) - 2(x-2) = 1(x-2)(x+3)
3x + 9 -2x + 4 = (x-2)(x+3)
x + 13 = x^2 + x - 6
x^2 - x + x - 6 - 13 = 0
x^2 - 19 = 0
x^2 = 19....take sq rt of both sides
sqrt(x^2) = sqrt 19
x = sqrt 19 and x = - sqrt 19 <====
I don’t know but I’m not gonna say anything so you don’t get it wrong
The fourth choice goes in the first box.
The second choices goes in the second box.
The fifth choice (the one on the bottom row) goes in the last box.
The fourth choice is 2/6. When reduced it equals 1/3.
The second choice is 3/6. When reduced it equals 1/2.
So when you add them, 2/6 + 3/6, you get 5/6.
Answer:
y=5x+2 goes to first table 1 + eq 6
2nd table goes to eq 4
3rd table goes to eq 5
4th table goes to eq 1
5th table goes to eq 2
6th table goes to eq 3
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