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laiz [17]
3 years ago
13

Need two examples using this chart. PLEASE!

Mathematics
1 answer:
Snowcat [4.5K]3 years ago
8 0

Answer:

We will use Descarte's Rule to solve the following question:

  • Work area to determine possible positive real roots:

It says that the number of sign changes in the function f(x) tells the maximum number of positive roots that could exist.

  • Work area to determine possible negative real roots:

Similarly the number of sign changes in the function f(-x) tells the maximum  number of negative roots that could exist.

Now let us consider a example as:

<h3>1) f(x)=x^3-x^2+5</h3>

We know that the total number of zeros of a polynomial function is always equal to the degree of the polynomial.

This is a polynomial function of degree 3 hence it has total 3 zeros.

Now f(x) has total 2 sign changes first from + to - and then from - to +.

Hence atmost 2 positive real zeros are possible.

Also f(-x)=-x^3-x^2+5

This function has only one sign change i.e. from - to +.

Hence atmost 1 negative real roots are possible.

Also on solving the cubic equation we got that we have one real zeros and 2 complex zeros.

In table we could write as:

<u>total zeros:</u>                 3    (-1.4334 ,  1.2167-1.4170 i ,  1.2167+1.4170 i)

<u>No. of positive : </u>          None

<u>real zero</u>

<u>No. of negative :</u>          1  (-1.4334)

<u>real zero</u>

<u>Complex zero :</u>              2  (1.2167-1.4170 i and  1.2167+1.4170 i)

<h3>2) f(x)=x^4+x^3-1</h3>

This is a polynomial function of degree 4 hence it has total 4 zeros.

Now f(x) has total 1 sign changes first from + to - .

Hence atmost 1 positive real zeros are possible.

Also f(-x)= x^4-x^3-1

This function has only one sign change i.e. from + to -.

Hence atmost 1 negative real roots are possible.

Also on solving the cubic equation we got that we have two real zeros and 2 complex zeros.

In table we could write as:

<u>total zeros:</u>                  4    (-1.3803, 0.81917 , -0.21945-0.91447 i ,    

                                                 ,   -0.21945+0.91447 i)    

<u>No. of positive : </u>          1 (0.81917)

<u>real zero</u>

<u>No. of negative :</u>          1  (-1.3803)

<u>real zero</u>

<u>Complex zero :</u>              2 (-0.21945-0.91447 i , -0.21945+0.91447 i)


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DanielleElmas [232]

Answer:

1

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that csc(x)=1/sin(x) and sec(x)=1/cos(x).

Thus:

\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}

=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }

Let's do the first part first: (Recall how to divide fractions)

\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}

For the second term:

\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}

So, all together: (same denominator; combine terms)

\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}

Note the numerator; it can be derived from the Pythagorean Identity:

sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)

Thus, we can substitute the numerator:

\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1

Everything simplifies to 1.

7 0
3 years ago
What quadrant is (25π/6) in
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Answer:

Not Sure

Step-by-step explanation:

The slope is 0.

See graph below...

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3 0
4 years ago
Without multiplying, tell which is greater: x 45 or x 45 explain.
Vanyuwa [196]

Answer: x45=x45 cannot be greater because they're both equivalent to each other.

Step-by-step explanation:

6 0
3 years ago
Question 6 (1.25 points)
Lisa [10]

Answer:

Test statistic Z= 1.713

The calculated Z- value =  1.7130 < 2.576 at 0.01 level of significance

Null hypothesis is accepted

There is no difference between the  mean annual salary of all lawyers in a city is  different from $110,000

Step-by-step explanation:

<u><em>Step(i):-</em></u>

A researcher wants to test if the mean annual salary of all lawyers in a city is

different from $110,000

Mean of the Population  μ = $110,000

Sample size 'n' = 53

Mean of the sample x⁻ = $114,000.

standard deviation of the Population = $17,000,

Level of significance = 0.01

Null hypothesis :

There is no difference between the  mean annual salary of all lawyers in a city is  different from $110,000

H₀:  x⁻ =  μ

Alternative Hypothesis :  x⁻ ≠  μ

<u><em>Step(ii)</em></u>:-

Test statistic

                 Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }

                 Z = \frac{114000-110000}{\frac{17000}{\sqrt{53} } }

                Z =  1.7130

Tabulated value Z = 2.576 at 0.01 level of significance

The calculated Z- value =  1.7130 < 2.576 at 0.01 level of significance

Null hypothesis is accepted

There is no difference between the  mean annual salary of all lawyers in a city is  different from $110,000

4 0
3 years ago
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