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seraphim [82]
3 years ago
11

Find the number. Round to the nearest tenth if necessary. 20% of 65 is what?

Mathematics
2 answers:
Alexxx [7]3 years ago
7 0

Answer: 13!

Step-by-step explanation : 0.20 x 65

Illusion [34]3 years ago
4 0
13 !! have a nice rest of your dayy
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Step-by-step explanation:

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Answer: The answer is C. 5

Step-by-step explanation:

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A class of 30 introductory statistics students took a quiz worth 100 points. the standard deviation of the scores was 0. what mu
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How do you solve this??? I need it fast please help with steps.<br> (x+3)(x+1)^2(x−4)&gt;0
scZoUnD [109]

Answer:

  x < -3 or x > 4

Step-by-step explanation:

The product of the factors is 4th-degree, and the leading coefficient is 1 (positive).

The factors tell you the following about the zeros:

  x = -3 . . . sign change from + to -

  x = -1 . . . graph touches, but no sign change. - on either side of x=-1

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The function will be positive for x < -3 and for x > +4.

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<em>Additional comments</em>

The value of the function is zero when any of its factors is zero. The value of a factor changes sign when the value of x changes from one side of the 0 to the other. For example, here, we have x+1 as a factor, so x=-1 is a zero. When x = -0.9, the factor is positive (-0.9 +1 = 0.1). When x = -1.1, the factor is negative (-1.1 +1 = -0.1). If this factor were to the first power, it would cause the value of the function to change sign at x=-1.

However, the factor (x+1) has an even power: (x+1)^2. That means when the factor is negative, its square is positive, and when the factor is positive, its square is positive. In short, the factor (x+1)^2 causes the function to be zero at x=-1, but does not make the function change sign there.

The product of all of these factors will result in a polynomial with x^4 as the highest-degree term. That means the function is of even degree. The leading coefficient is 1, so is positive, and the function will generally have a U-shape. The left-most (odd-degree) zero will be where the function changes sign from positive to negative. The right-most (odd-degree) zero will be where the function changes sign from negative to positive. Those zeros are x=-3 and x=+4, respectively. There are no places between these where the function changes sign, so these zeros are the ends of the regions where the function is positive (>0).

3 0
2 years ago
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustra
Rainbow [258]

Answer:

x = t

y = 1 - t

z = 2t

Step-by-step explanation:

Given

x=t

y=e^{-t}

z=2t-t^2

(0, 1, 0)

The vector equation is given as:

r(t) = (x,y,z)

Substitute values for x, y and z

r(t) = (t,\ e^{-t},\ 2t - t^2)

Differentiate:

r'(t) = (1,\ -e^{-t},\ 2 - 2t)

The parametric value that corresponds to (0, 1, 0) is:

t = 0

Substitute 0 for t in r'(t)

r'(t) = (1,\ -e^{-t},\ 2 - 2t)

r'(0) = (1,\ -e^{-0},\ 2 - 2*0)

r'(0) = (1,\ -1,\ 2 - 0)

r'(0) = (1,\ -1,\ 2)

The tangent line passes through (0, 1, 0) and the tangent line is parallel to r'(0)

It should be noted that:

The equation of a line through position vector a and parallel to vector v is given as:

r(t) = a + tv

Such that:

a = (0,1,0) and v = r'(0) = (1,-1,2)

The equation becomes:

r(t) = (0,1,0) + t(1,-1,2)

r(t) = (0,1,0) + (t,-t,2t)

r(t) = (0+t,1-t,0+2t)

r(t) = (t,1-t,2t)

By comparison:

r(t) = (x,y,z) and r(t) = (t,1-t,2t)

The parametric equations for the tangent line are:

x = t

y = 1 - t

z = 2t

7 0
2 years ago
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