The cube root values and a graph of them are shown in the attachment.
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The cube root of a negative number is negative. These all have exact (rational) cube roots.
A long wire is attached to the top of a 175 foot tall electrical transmission tower. If the wire needs to an angle of 65 degrees
1 answer:
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Answer:
<u>0.9524</u>
Step-by-step explanation:
<em>Note enough information is given in this problem. I will do a similar problem like this. The problem is:</em>
<em>The Probability of a train arriving on time and leaving on time is 0.8.The probability of the same train arriving on time is 0.84. The probability of the same train leaving on time is 0.86.Given the train arrived on time, what is the probability it will leave on time?</em>
<em />
<u>Solution:</u>
This is conditional probability.
Given:
- Probability train arrive on time and leave on time = 0.8
- Probability train arrive on time = 0.84
- Probability train leave on time = 0.86
Now, according to conditional probability formula, we can write:
= P(arrive ∩ leave) / P(arrive)
Arrive ∩ leave means probability of arriving AND leaving on time, that is given as "0.8"
and
P(arrive) means probability arriving on time given as 0.84, so:
0.8/0.84 = <u>0.9524</u>
<u></u>
<u>This is the answer.</u>
<h3>Answer:</h3>
- f(x) = -2x^3 +3x^2 +11x -6
- see attached
- an infinite number. Since the magnitude of the leading coefficient is not specified, it may be any negative number. (We have chosen the smallest magnitude integer that makes all coefficients be integers.)
1. When "a" is a root of a polynomial, (x -a) is a factor of it. For the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).
In order to make the leading coefficient be negative, we need to multiply this product by a negative number. Any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.
Then ...
... f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)
... = -(2x² +3x -2)(x -3)
... = -(2x³ -3x² -11x +6)
... f(x) = -2x³ +3x² +11x -6
2. A graph created by the Desmos on-line graphing calculator is shown, and the zeros are highlighted.
3. As indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. You want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.