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garri49 [273]
1 year ago
6

What is the quotient of (2x4 – 3x3 – 3x2 7x – 3) ÷ (x2 – 2x 1)? 2 x superscript 4 baseline minus 3 x cubed minus eleven-halves 2

x2 x – 3 – startfraction 6 over x squared minus 2 x 1 endfraction 2x2 – x 2x2 x – 3
Mathematics
1 answer:
evablogger [386]1 year ago
7 0

The answer choice which represents the quotient of the polynomials given is; 2x² +x -3.

<h3>What is the quotient of the polynomial division?</h3>

According to the task content, the quotient of the polynomial division; (2x4 – 3x3 – 3x2 7x – 3) ÷ (x2 – 2x 1) is required;

Hence, it follows from long division of polynomials that the required quotient is; 2x² +x -3.

Read more on quotients of polynomials;

brainly.com/question/24662212

#SPJ4

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Solve x2 − 12x + 5 = 0 using the completing-the-square method. x = six plus or minus the square root of five x = negative six pl
Andru [333]

we are given

x^2-12x+5=0

we have to solve it by completing square method

step-1: Move 5 on right side

x^2-12x+5-5=0-5

x^2-12x=-5

step-2: Break middle term

x^2-2*6*x=-5

step-3: Add 6^2 both sides

x^2-2*6*x+(6)^2=-5+(6)^2

(x-6)^2=31

step-3: Solve for x

\sqrt{(x-6)^2} =\sqrt{31}

we wil get two values

First value is

x-6=\sqrt{31}

add both sides 6

x-6+6=6+\sqrt{31}

x=6+\sqrt{31}

Second value is

x-6=-\sqrt{31}

add both sides 6

x-6+6=6-\sqrt{31}

x=6-\sqrt{31}

so, solutions are

x=6+\sqrt{31}

x=6-\sqrt{31}.............Answer

5 0
3 years ago
In a test of the effectiveness of garlic for lowering​ cholesterol, 8181 subjects were treated with raw garlic. Cholesterol leve
xz_007 [3.2K]

Answer:

With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

Step-by-step explanation:

The dependent <em>t</em>-test (also known as the paired <em>t</em>-test or paired samples <em>t</em>-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired <em>t</em>-test is used to determine the effectiveness of garlic for lowering​ cholesterol.

A random sample of 81 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment.

The hypothesis for the test can be defined as follows:

<em>H₀</em>: With garlic​ treatment, the mean change in LDL cholesterol is not greater than 0, i.e. <em>d</em> ≤ 0.

<em>Hₐ</em>: With garlic​ treatment, the mean change in LDL cholesterol is greater than 0, i.e. <em>d</em> > 0.

The information provided is:

\bar d=0.40\\SD_{d}=16.2\\\alpha =0.01

Compute the test statistic value as follows:

t=\frac{\bar d}{SD_{d}/\sqrt{n}}\\\\=\frac{0.40}{16.2/\sqrt{81}}\\\\=0.22

The test statistic value is 0.22.

Decision rule:

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Compute the <em>p</em>-value of the test as follows:

p-value=P(t_{n-1}>0.22)\\=P(t_{80}>0.22)\\=0.4132

*Use a <em>t</em>-table.

The <em>p</em>-value of the test is 0.4132.

<em>p-</em>value= 0.4132 > <em>α</em> = 0.01

The null hypothesis was failed to be rejected.

Thus, it can be concluded that with garlic​ treatment, the mean change in LDL cholesterol is not greater than 0.

3 0
2 years ago
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