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Nookie1986 [14]
3 years ago
14

The temperature falls from 0 degrees to -125 degrees in 3 hours. Which expression finds the change in . temperature per hour? 07

:49 7 49 2 4 7 49 0 -3*4 49 4 x 49 7 4 -​
Mathematics
1 answer:
bixtya [17]3 years ago
6 0

Answer:

-3 *4 449 expression finds the change in

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Find all the zeros of the equation x^4-6x^2-7x-6=0 Explain please.
Alina [70]
<h3>Answer:</h3>
  • zeros are {-2, 3, (-1±i√3)/2}
<h3>Step-by-step explanation:</h3>

I like to look at a graph of the function to see where the zeros might be. Here, there are x-intercepts at x=-2 and x=3. These can be factored out using synthetic division to find the factorization to be ...

... (x +2)(x -3)(x² +x +1) = 0

By completing the square, using the quadratic formula, or by looking at the graph of it, the complex roots of the quadratic factor can be found to be ...

... x = (-1 ±i√3)/2

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The second attachment shows my synthetic division. The first division takes out the root x=3 to give a quotient of x³ +3x² +3x +2. The second division takes out the root -2 to give the quotient of x² +x +1. (You can see that I tried -1 as a root first.)

The graph shows both the quartic and the quadratic factor of it. The latter has a leading coefficient of 1 and a vertex at (-1/2, 3/4), so you know the complex roots are -1/2 ±i√(3/4).

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<em>From the beginning</em>

There is only a very complicated formula for the roots of a quartic equation, so these are usually solved by machine or by some form of trial and error (iteration). There are some helps, like Descarte's Rule of Signs, and the Rational Root theorem.

Here, the former looks at the one sign change in the coefficients to tell you there will be 1 positive real root. Changing the sign of the odd-degree terms makes there be 3 sign changes, so there will be 3 or 1 negative real roots. Thus, we're assured at least two real roots, one of each sign.

We can look at the constant term to find the y-intercept to be -6. We can add the coefficients to find the value of the function is -18 for x=1, so the positive real root is larger than 1.

The Rational Root theorem says any rational roots will be factors of 6, the constant term. Choices are 1, 2, 3, 6. We have already eliminated 1 as a possibility, and we consider it unlikely that 6 will be a root. (The 4th power overwhelms the other terms in the function.) We tried 2 and found it doesn't work (this was before we graphed the function). The attached division result shows that 3 is a root, as does the graph.

Once you get down to a quadratic, you can find the remaining roots in the usual way. Because it is so simple to read them from the graph, we decided to graph the quadratic factor.

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<em>Comment on terminology</em>

"root" and "zero" are essentially the same thing when the function is equated to zero, as here. The terms refer to the value(s) of x that make the polynomial function evaluate to zero.

8 0
3 years ago
Suppose that the hourly wages of fast food workers are normally distributed with an unknown mean and standard deviation. The wag
Nutka1998 [239]

Answer: 1.303639

Step-by-step explanation:

The t-score for a level of confidence (1-\alpha) is given by :_

t_{(df,\alpha/2)}, where df is the degree of freedom and \alpha is the significance level.

Given : Level of significance : 1-\alpha:0.80

Then , significance level : \alpha: 1-0.80=0.20

Sample size : n=40

Then , the degree of freedom for t-distribution: df=n-1=40-1=39

Using the normal t-distribution table, we have

t_{(df,\alpha/2)}=t_{39,0.10}=1.303639

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8 0
3 years ago
How do I round off to the nearest decimal place​
Marrrta [24]
First, find the digit in the place you are rounding to. Look at the digit one place to the right. If the digit is less than 5, round down. If the digit is 5 or greater, round up.
For example: round 1.86 to the nearest tenth
Look at the tenth digit (8). The number to the right of it is 6. Since it’s above 5, round the 8 up to 9. So you would get 1.9.

If it was 1.84, since 4 is less than 5, you would not round the 8, and you would get 1.8
3 0
3 years ago
What is the cross-sectional area of a cement pipe 2 in. thick with an inner diameter of 5 ft? Use pie 3.14.
Paul [167]

Answer:

1.60 ft^{2}

Step-by-step explanation:

Assuming that this is a cylindrical pipe, the area can be determined by applying the formula for calculating the area of a circle.

i.e Area = \pir^{2}

So that,

for the inner part of the pipe,

r = \frac{diameter}{2}

 = \frac{5}{2}

 = 2.5 feet

Area of the inner part of the pipe = \pir^{2}

                                 = 3.14 x (2.5)^{2}

                                 = 19.625

Are of the inner part of the pipe is 19.63 ft^{2}.

Total diameter of the pipe = 5 + 0.2

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r = \frac{5.2}{2}

 = 2.6 feet

Area of the pipe = \pir^{2}

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                            = 21.2264

Area of the pipe is 21.23 ft^{2}.

Thus the cross sectional area = Area of the pipe - Area of the inner part of the pipe

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The cross sectional area of the cement pipe is 1.60 ft^{2}.

6 0
2 years ago
Write a verbal expression for 4n^3+6<br><br> Please Help!
Vlad [161]

The verbal expression is "<em>Four enn-cubed plus six</em>."
7 0
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