3. Emma said that when you multiply three negative decimals , the product will be positive . Use these numbers to answer the que
1 answer:
Using multiplication signal rules, it is found that:
A: Emma's statement is always false.
B: The result is always negative.
C: Emma's statement is always true.
The rule used for this exercise is as follows:
- When two numbers of different signals are multiplied, the result is negative.
- When two numbers have the same signal, the result is positive.
Part A:
- Three numbers are multiplied, all negative.
- The multiplication of the first two result in a positive number.
- Then, this positive number is multiplied by a negative number, and the result will be negative, which mean that Emma's statement is always false.
Two examples are:
Part B:
The rule is that the result is always negative.
Part C:
- The multiplication of the first two negative numbers result in a positive number.
- Then, this positive number is multiplied by another positive number, and the result will be positive, which mean that Emma's statement is always true.
Two examples are:
A similar problem is given at brainly.com/question/24764960
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Answer:
f(r) = (x -1)(x -4+√7)(x -4-√7)
Step-by-step explanation:
The signs of the terms are + - + -. There are 3 changes in sign, so Descartes' rule of signs tells you there are 3 or 1 positive real roots.
The rational roots, if any, will be factors of 9, the constant term. The sum of coefficients is 1 -9 +17 -9 = 0, so you know that r=1 is one solution to f(r) = 0. That means (r -1) is a factor of the function.
Using polynomial long division, synthetic division (2nd attachment), or other means, you can find the remaining quadratic factor to be r^2 -8r +9. The roots of this can be found by various means, including completing the square:
r^2 -8r +9 = (r^2 -8r +16) +9 -16 = (r -4)^2 -7
This is zero when ...
(r -4)^2 = 7
r -4 = ±√7
r = 4±√7
Now, we know the zeros are {1, 4+√7, 4-√7), so we can write the linear factorization as ...
f(r) = (r -1)(r -4 -√7)(r -4 +√7)
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<em>Comment on the graph</em>
I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)
