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8090 [49]
2 years ago
14

If f(x)=4(3x-5), find f^-1(x)

Mathematics
1 answer:
Margaret [11]2 years ago
3 0

Answer:

f^-1(x) = (x+20) / 12

Step-by-step explanation:

f(x) = 4(3x-5)

Let y be the image of f.

y = 4(3x-5)

y = 12x-20

y+20 = 12x

x = (y+20) / 12

f^-1(y) = (y+20) / 12, so

f^-1(x) = (x+20) / 12

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disa [49]

Answer:

d = 3

Step-by-step explanation:

6*3=18
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The classes At the middle school want to rise money the 6 grade runs a bake sale 5 hours and makes 170 the 7 grade sets up a dru
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Sixth grade
Because if you divide their total profit by the total amount of hours they worked, you would find that the sixth grade's income per hour is the highest.
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8 0
3 years ago
What is the equation for the plane illustrated below?
TiliK225 [7]

Answer:

Hence, none of the options presented are valid. The plane is represented by 3 \cdot x + 3\cdot y + 2\cdot z = 6.

Step-by-step explanation:

The general equation in rectangular form for a 3-dimension plane is represented by:

a\cdot x + b\cdot y + c\cdot z = d

Where:

x, y, z - Orthogonal inputs.

a, b, c, d - Plane constants.

The plane presented in the figure contains the following three points: (2, 0, 0),  (0, 2, 0), (0, 0, 3)

For the determination of the resultant equation, three equations of line in three distinct planes orthogonal to each other. That is, expressions for the xy, yz and xz-planes with the resource of the general equation of the line:

xy-plane (2, 0, 0) and (0, 2, 0)

y = m\cdot x + b

m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Where:

m - Slope, dimensionless.

x_{1}, x_{2} - Initial and final values for the independent variable, dimensionless.

y_{1}, y_{2} - Initial and final values for the dependent variable, dimensionless.

b - x-Intercept, dimensionless.

If x_{1} = 2, y_{1} = 0, x_{2} = 0 and y_{2} = 2, then:

Slope

m = \frac{2-0}{0-2}

m = -1

x-Intercept

b = y_{1} - m\cdot x_{1}

b = 0 -(-1)\cdot (2)

b = 2

The equation of the line in the xy-plane is y = -x+2 or x + y = 2, which is equivalent to 3\cdot x + 3\cdot y = 6.

yz-plane (0, 2, 0) and (0, 0, 3)

z = m\cdot y + b

m = \frac{z_{2}-z_{1}}{y_{2}-y_{1}}

Where:

m - Slope, dimensionless.

y_{1}, y_{2} - Initial and final values for the independent variable, dimensionless.

z_{1}, z_{2} - Initial and final values for the dependent variable, dimensionless.

b - y-Intercept, dimensionless.

If y_{1} = 2, z_{1} = 0, y_{2} = 0 and z_{2} = 3, then:

Slope

m = \frac{3-0}{0-2}

m = -\frac{3}{2}

y-Intercept

b = z_{1} - m\cdot y_{1}

b = 0 -\left(-\frac{3}{2} \right)\cdot (2)

b = 3

The equation of the line in the yz-plane is z = -\frac{3}{2}\cdot y+3 or 3\cdot y + 2\cdot z = 6.

xz-plane (2, 0, 0) and (0, 0, 3)

z = m\cdot x + b

m = \frac{z_{2}-z_{1}}{x_{2}-x_{1}}

Where:

m - Slope, dimensionless.

x_{1}, x_{2} - Initial and final values for the independent variable, dimensionless.

z_{1}, z_{2} - Initial and final values for the dependent variable, dimensionless.

b - z-Intercept, dimensionless.

If x_{1} = 2, z_{1} = 0, x_{2} = 0 and z_{2} = 3, then:

Slope

m = \frac{3-0}{0-2}

m = -\frac{3}{2}

x-Intercept

b = z_{1} - m\cdot x_{1}

b = 0 -\left(-\frac{3}{2} \right)\cdot (2)

b = 3

The equation of the line in the xz-plane is z = -\frac{3}{2}\cdot x+3 or 3\cdot x + 2\cdot z = 6

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a = 3, b = 3, c = 2, d = 6

Hence, none of the options presented are valid. The plane is represented by 3 \cdot x + 3\cdot y + 2\cdot z = 6.

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IgorLugansk [536]

Answer:

1/1000

Step-by-step explanation:

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