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2 years ago

4. Which of the following can be used

1 answer:

7 0

The option that can be used to support the idea that the set of polynomials is closed under multiplication is; Option C: (10x^(0.5) - 8)(5x^(0.5) + 4)

<h3>What is the Closure property under multiplication?</h3>

When multiplying polynomials, the variables' exponents are added, according to the rules of exponents. It is pertinent to note that the exponents in polynomials are whole numbers. The whole numbers are closed under addition, which guarantees that the new exponents will be whole numbers. Thus, we can also say that the polynomials are closed under multiplication.

Now, looking at the options, we can say that option C is the only polynomial that is closed under multiplication because its' variables and exponents will not change;

(10x^(0.5) - 8)(5x^(0.5) + 4)

The output will retain the same thing.

Read more about closure property at; brainly.com/question/19340450

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3 years ago

The data below show the weights, in pounds, of a group of professional football players.

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Answer:

The Mean Absolute Deviation is 11

Step-by-step explanation:

Given:

325, 310, 289, 288, 285, 285, 285, 280, 280 and 273

Mean = 290

Required:

Calculate the Mean Absolute Deviation

Provided that we have the value of the mean to be 290 from the question; the following steps will calculate the Mean Absolute Deviation

Step 1: Subtract the mean weight from each weight

325 - 290 = 35

310 - 290 = 20

289 - 290 = -1

288 - 290 = -2

285 - 290 = -5

280 - 290 = -10

273 - 290 = -17

Step 2: Calculate the absolute value of the each results above

|35| = 35

|20| = 20

|-1| = 1

|-2| = 2

|-5| = 5

|-10| = 10

|-17| = 17

Step 3: Calculate the mean of the data above

Mean Absolute Value = $\frac{35 + 20 + 1 + 2 + 5 + 5 + 5 + 10 + 10 + 17}{10}$

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3 years ago

What is the discontinuity and zero of the function f(x) = 3x^2 + x - 4 / x-1

7nadin3 [17]

Ans: Option (1) <span>Discontinuity at (1, 7), zero at ( negative four thirds , 0)
</span>
Explanation:
Given function:
$f(x) = \frac{3x^2 + x - 4}{x-1}$

Now if we plug in the x = 1, we would have discontinuity as function goes to infinity.

Now for f(1):
$f(x) = \frac{3x^2 + x - 4}{x-1} \\ f(x) = \frac{3x^2 + 4x - 3x - 4}{x-1} \\ f(x) = \frac{x(3x+4)-1(3x+4)}{x-1} \\ f(x) = \frac{(x-1)(3x+4)}{(x-1)} \\ f(x) = 3x+4 \\ now ~ insert ~ x=1: \\ f(1) = 3(1) + 4 = 7 \\ It~means~discontinuity~at~(1,7).\\ Now~let~us~find~zeros.\\ put~ f(x) = 3x+4 ~ equals~ to ~zero. \\ =\ \textgreater \ ~3x+4 = 0 \\ =\ \textgreater \ ~ x = \frac{-4}{3} \\ Hence~zeros=( \frac{-4}{3}, 0)$

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3 years ago