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2 years ago

Without using a calculator, determine the number of real zeros of the function

Mathematics

1 answer:

kumpel [21]2 years ago

3 0

The number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3

<h3>How to determine the number of real zeros?</h3>

The equation of the function is given as:

$f(x) = x^3 + 4x^2 + x - 6$

Expand the function

$f(x) = x^3 + 5x^2 - x^2 + 6x - 5x - 6$

Reorder the terms

$f(x) = x^3 + 5x^2 + 6x - x^2 - 5x - 6$

Factor the expression

$f(x) = x(x^2 + 5x + 6) -1(x^2 + 5x + 6)$

Factor out x -1

$f(x) = (x^2 + 5x + 6)(x -1)$

Expand

$f(x) = (x^2 + 3x + 2x + 6)(x -1)$

Factorize

$f(x) = [x(x + 3) + 2(x + 3)](x - 1)$

Factor out x + 2

$f(x) = (x + 3)(x + 2)(x- 1)$

The function has been completely factored and it has 3 linear factors

Hence, the number of real zeros of the function f(x) = x3 + 4x2 + x − 6 is 3

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Answer:

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Step-by-step explanation:

Here we have to look at the given graph and find the answer.

The vertical axis represents the price and cost per unit. That is represented by P.

The horizontal axis represents the units of output, that is represented by q.

We need to find the number of units when p = $55.

By look at the graph of the curve MC, when P = $55, the number of units is 50.

Therefore, the answer is C. 50.

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3 years ago

Read 2 more answers

In a given year there is a 3% chance you will get in a small accident costing around $2000 and 0.1% chance you will get in a maj

Wittaler [7]

Answer:

$475

Step-by-step explanation:

There are 3 possible accident in this question

3% chance of losing $2000

0.1% chance of losing $150,000

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Then the expected value that you will lose is:

3%* $2000 + (0.1% * $15000) + (96.9% * $0)= $75

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4 years ago

Please help me with my test( will give you brainiest to whoever does it)

olasank [31]

Answer:

all i can do

Step-by-step explanation:

2. if you roll a six-sided dice one time and flip a coin one time what is the probability that you flip an odd number on a die and heads on a die.

3. The probability would be 1/2 or 0.5.

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3 years ago

Use the binomial expression (p+q)^n to calculate a binomial distribution with n=5 and p=0.3.(Show all steps)

Helen [10]

Answer:

The binomial in expanded form is $(0.3 + q)^{5} = \frac{243}{100000} + \frac{81}{2000}\cdot q + \frac{27}{100}\cdot q^{2} + \frac{9}{10} \cdot q^{3} + \frac{3}{2}\cdot q^{4} + q^{5}$ .

Step-by-step explanation:

The Binomial Theorem states that a binomial of the form $(a + b)^{n}$ can be expanded by using the following identity:

$(a + b)^{n} = \Sigma \limits^{n}_{k = 0}\,\frac{n!}{k!\cdot (n-k)!}\cdot a^{n-k}\cdot b^{k}$ (1)

If we know that $a = p = 0.3$ and $n = 5$ , then the expanded form of the binomial is:

$(p+q)^{n} = \frac{243}{100000} + 5\cdot \left(\frac{81}{10000} \right)\cdot q + 10\cdot \left(\frac{27}{1000})\cdot q^{2} + 10\cdot \left(\frac{9}{100} \right)\cdot q^{3} + 5\cdot \left(\frac{3}{10} \right)\cdot q^{4} + q^{5}$

$(0.3 + q)^{5} = \frac{243}{100000} + \frac{81}{2000}\cdot q + \frac{27}{100}\cdot q^{2} + \frac{9}{10} \cdot q^{3} + \frac{3}{2}\cdot q^{4} + q^{5}$

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3 years ago