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4 years ago

For the function f(x)=2x^3-3x^2-22.5x+19

1 answer:

7 0

A. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5\\\\f''(x)=12x-6\\12x-6=0\\12x=6\\x=0.5$ - Relative maximum at x = -1.5. Relative minimum at x = 2.5. Point of inflection at x = 0.5.

B. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5$ - Absolute maximum at x = -1.5. Absolute minimum at x = 2.5. This applies for the interval [-4, 5].

C. $f(x)=2x^3-3x^2-22.5x+19\\f(0)=2(0)^3-3(0)^2-22.5(0)+19\\f(0)=19\\\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=2.5$ - Absolute maximum at x = 0. Absolute minimum at x = 2.5. This applies for the interval [0, 3].

D. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5$ - Test: $f(-2)=2x^3-3x^2-22.5x+19\\f(-2)=2(-2)^3-3(-2)^2-22.5(-2)+19\\f(-2)=2(-8)-3(4)-45+19\\f(-2)=-16-12-26\\f(-2)=-540\\\\f(3)=2x^3-3x^2-22.5x+19\\f(3)=2(3)^3-3(3)^2-22.5(3)+19\\f(3)=2(27)-3(9)-67.5+19\\f(3)=54-27-48.5\\f(3)=-21.5$ - Increasing: $[-\infty,-1.5)(2.5,\infty]$ - Decreasing: $(-1.5,2.5)$

E. $f(x)=2x^3-3x^2-22.5x+19\\f'(x)=6x^2-6x-22.5\\f'(x)=6x^2-6x-22.5=0\\4x^2-4x-15=0\\(2x+3)(2x-5)=0\\2x+3=0~|~2x-5=0\\2x=-3~|~2x=5\\x=-1.5~|~x=2.5\\x=-1.5,2.5\\\\f''(x)=12x-6\\12x-6=0\\12x=6\\x=0.5$ - Concave up: $(0.5,\infty]$ - Concave down: $[-\infty,0.5)$

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Answer:

Simplifying

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6x + -2y = 30

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '2y' to each side of the equation.

6x + -2y + 2y = 30 + 2y

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Step-by-step explanation:

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3 years ago

A set of data has a mean of 45.6 what is the mean if 5.0 is added to each score

Solnce55 [7]

Answer:

The mean will be increased by 5

Step-by-step explanation:

Suppose a set of data $(2, 4, 6, 8, 10, 12)$

Mean is defined as sum of all the values given set of data divided by total number of values.

Mean1 = $\frac{2+4+6+8+10+12}{6} = \frac{42}{6} = 7$

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3 years ago

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Yanka [14]

Answer:

$x=\frac{12}{7} \\y=\frac{12}{5} \\z=-12$

Step-by-step explanation:

Let's re-write the equations in order to get the variables as separated in independent terms as possible \:

First equation:

$\frac{xy}{x+y} =1\\xy=x+y\\1=\frac{x+y}{xy} \\1=\frac{1}{y} +\frac{1}{x}$

Second equation:

$\frac{xz}{x+z} =2\\xz=2\,(x+z)\\\frac{1}{2} =\frac{x+z}{xz} \\\frac{1}{2} =\frac{1}{z} +\frac{1}{x}$

Third equation:

$\frac{yz}{y+z} =3\\yz=3\,(y+z)\\\frac{1}{3} =\frac{y+z}{yz} \\\frac{1}{3}=\frac{1}{z} +\frac{1}{y}$

Now let's subtract term by term the reduced equation 3 from the reduced equation 1 in order to eliminate the term that contains "y":

$1=\frac{1}{y} +\frac{1}{x} \\-\\\frac{1}{3} =\frac{1}{z} +\frac{1}{y}\\\frac{2}{3} =\frac{1}{x} -\frac{1}{z}$

Combine this last expression term by term with the reduced equation 2, and solve for "x" :

$\frac{2}{3} =\frac{1}{x} -\frac{1}{z} \\+\\\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\ \\\frac{7}{6} =\frac{2}{x}\\ \\x=\frac{12}{7}$

Now we use this value for "x" back in equation 1 to solve for "y":

$1=\frac{1}{y} +\frac{1}{x} \\1=\frac{1}{y} +\frac{7}{12}\\1-\frac{7}{12}=\frac{1}{y} \\ \\\frac{1}{y} =\frac{5}{12} \\y=\frac{12}{5}$

And finally we solve for the third unknown "z":

$\frac{1}{2} =\frac{1}{z} +\frac{1}{x} \\\\\frac{1}{2} =\frac{1}{z} +\frac{7}{12} \\\\\frac{1}{z} =\frac{1}{2}-\frac{7}{12} \\\\\frac{1}{z} =-\frac{1}{12}\\z=-12$

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Answer:

B, C

Step-by-step explanation:

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