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3 years ago

What is the inverse of the function f(x) = 8x+1

Mathematics

1 answer:

Yanka [14]3 years ago

6 0

Answer:

f⁻¹(x) = (x - 1)/8

f⁻¹(x) = 1/8 x - 1/8

Step-by-step explanation:

To find the inverse of a function, switch the "x" and "y" variables, then isolate "y".

Remember "f(x)" is the same thing as "y". Change from function notation to "y".

f(x) = 8x + 1

y = 8x + 1

Switch the "x" and "y" variables

x = 8y + 1

Isolate "y". Move the "y" variable to the left for standard formatting

8y + 1 = x

8y + 1 - 1 = x - 1 Subtract 1 from both sides

8y = x - 1

$\frac{8y}{8}=\frac{x-1}{8}$ Divide both sides by 8 and simplify

$y=\frac{x-1}{8}$ Inverse equation

$y=\frac{1}{8}x-\frac{1}{8}$ Slope-intercept form

Use function notation, change "y"

$f^{-1}(x)=\frac{x-1}{8}$ Simplified

$f^{-1}(x)=\frac{1}{8}x-\frac{1}{8}$ Slope-intercept form

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Step-by-step explanation:

2/3 dress in 4 hours

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2 years ago

A right triangle has one angle that measures 6 degrees. What is the measure of the other acute angle?

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Step-by-step explanation:

All three angles add up to 180 degrees.

Since it is a right triangle, there is a 90 degrees angle. And we know another angle is 6 degree. So we use the formula: 180 - (known angle + known angle) = unknown angle

Plugging in the information we get: 180 - (90 + 6) = x

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2 years ago

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Answer:

Linearly Dependent for not all scalars are null.

Step-by-step explanation:

Hi there!

1)When we have vectors like $v_{1},v_{2},v_{3}, ...$ we call them linearly dependent if we have scalars $a_{1},a_{2},a_{3},...$ as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

$a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+a_{3}\vec{v_{3}}+...a_{m}\vec{v_{m}}=0$

When all scalar coefficients are equal to zero, we can call them linearly independent

2) Now let's examine the Matrix given:

$\begin{bmatrix}1 &-2 &2 &3 \\ -2 & 4 & -4 &3 \\ 0&1 &-1 & 4\end{bmatrix}$

So each column of this Matrix is a vector. So we can write them as:

$\vec{v_{1}}=\left \langle 1,-2,1 \right \rangle,\vec{v_{2}}=\left \langle -2,4,-1 \right \rangle,\vec{v_{3}}=\left \langle 2,-4,4 \right \rangle\vec{v_{4}}=\left \langle 3,3,4 \right \rangle$ Or

Now let's rewrite it as a system of equations:

$a_{1}\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_{2}\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_{3}\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_{4}\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

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