All categories

3 years ago

How do u solve this equation?

Mathematics

2 answers:

Mnenie [13.5K]3 years ago

7 0

Answer:

x^3+4x^2+3x-4 R 9

Step-by-step explanation:

There are two ways to solve polynomial division problems, long division, and synthesis. Synthesis won't work because the divisor is not in ax+c form, where a is always 1, and c is any integer. So, long division must be used. I would work it out here, but that would take for ever, so I provided a link to a website that shows you how to solve this.

Hope this helps! :)

MrRa [10]3 years ago

6 0

You have to do a thing called PEMDAS

You might be interested in

Which equation creates an infinite number of solutions when solved for a system with <img src="https://tex.z-dn.net/?f=y%3D8x-9"

Tatiana [17]

Answer:

d) 4y − 32x = -36

Step-by-step explanation:

If there are an infinite number of solutions, the equations represent the same line.

y = 8x − 9

y − 8x = -9

4y − 32x = -36

6 0

3 years ago

Read 2 more answers

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the

Alexxandr [17]

Answer:

Step-by-step explanation:

The table can be computed as:

Advertising Expenses ($ million) Number of companies

25 up to 35 4

35 up to 45 19

45 up to 55 27

55 up to 65 16

65 up to 75 9

TOTAL 75

Let's find the probabilities first:

$P(25 - 35) = P \Big(\dfrac{25-50.93}{10.80}$

For 35 up to 45

$P(35 - 45) = P \Big(\dfrac{35-50.93}{10.80}$

For 45 up to 55

$P(45 - 55) = P \Big(\dfrac{45-50.93}{10.80}$

For 55 up to 65

$P(55 - 65) = P \Big(\dfrac{55-50.93}{10.80}$

For 65 up to 75

$P(65 - 75) = P \Big(\dfrac{65-50.93}{10.80}$

Chi-Square Table can be computed as follows:

Expense No of Probabilities(P) Expe $(O-E)^2$ $\dfrac{(O-E)^2}{E}$

compa cted E (n*p)

nies (O)

25-35 4 0.0612 75*0.0612 = 4.59 0.3481 0.0758

35-45 19 0.2218 75*0.2218 = 16.635 5.5932 0.3362

45-55 27 0.3568 75*0.3568 = 26.76 0.0576 0.021

55-65 16 0.2552 75*0.2552 = 19.14 9.8596 0.5151

65-75 9 0.0839 75*0.0839 = 6.2925 7.331 1.1650

$\sum \dfrac{(O-E)^2}{E}= 2.0492$

Using the Chi-square formula:

$X^2 = \dfrac{(O-E)^2}{E} \\ \\ Chi-square \ X^2 = 2.0942$

Null hypothesis:

$H_o: \text{The population of advertising expenses follows a normal distribution}$

Alternative hypothesis:

$H_a: \text{The population of advertising expenses does not follows a normal distribution}$

Assume that:

$\alpha = 0.02$

degree of freedom:

= n-1

= 5 -1

= 4

Critical value from $X^2 = 11.667$

Decision rule: To reject $H_o \ if \ X^2$ test statistics is greater than $X^2$ tabulated.

Conclusion: Since $X^2 = 2.0942$ is less than critical value 11.667. Then we fail to reject $H_o$

6 0

3 years ago

Help me, please!! Estoy desesperada y no hablo inglês!! Help

Aleksandr-060686 [28]

$D:x>0\\
\log x+\log x^2+\log x^3=-6\\
\log (x\cdot x^2\cdot x^3)=\log 10^{-6}\\
x^6=10^{-6}\\
x^6=\left(\frac{1}{10}\right)^6\\ \boxed{x=\frac{1}{10}}$