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Sergeu [11.5K]
3 years ago
5

Need help on this math problem!!!

Mathematics
1 answer:
Colt1911 [192]3 years ago
4 0

Answer:

(fof^{-1})(x)=x

Step-by-step explanation:

Composition of two functions f(x) and g(x) is represented by,

(fog)(x) = f[g(x)]

If a function is,

f(x) = (-6x - 8)² [where x ≤ -\frac{8}{6}]

Another function is the inverse of f(x),

f^{-1}(x)=-\frac{\sqrt{x}+8}{6}

Now composite function of these functions will be,

(fof^{-1})(x)=f[f^{-1}(x)]

                  = [-6(\frac{\sqrt{x}+8}{6})-8]^{2}

                  = [-\sqrt{x}+8-8]^2

                  = (-\sqrt{x})^2

                  = x

Therefore, (fof^{-1})(x)=x

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If the water level in a lake drops by 14 feet, the water level will still be higher than the 20-foot mark that triggers restrict
maria [59]

Answer:

A is correct

Step-by-step explanation:

Here, we want to write an inequality

If the water level drops by 14 feet

what this mean is that we are subtracting 14 from the original level L

Thus, we have

L-14

Higher than 20 means that we have;

L-14 > 20

L > 20 + 14

L > 34

8 0
3 years ago
Name: Co
White raven [17]

Answer:

Thus last month Calvin made $414.90 from all the paintings sold at the art shows.

Step-by-step explanation:

Let us first analyse the given information in the question:

→ Small paints (lets call them s) sell for $25.80 Per Painting, thus:

  25.80s gives as the profit for s number paints sold.

→ Large paints (lets call them l) sell for $56.25 Per Painting, thus:

  56.25l gives as the profit for l number paints sold.

We also know last month he sold six large paintings and three small paintings thus from our variables above we can say:

s=3\\l=6         Eqn(1).

Thus the total profit of all paintings sold would be:

PROFIT= 25.80s+56.25l       which by pluggin in Eqn(1) values gives

PROFIT=25.80*3+56.25*6\\PROFIT=414.90

Thus last month Calvin made $414.90 from all the paintings sold at the art shows.

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4 years ago
For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.
Anni [7]

Answer:

The reduced row-echelon form of the linear system is \left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

  1. Interchange two rows
  2. Multiply one row by a nonzero number
  3. Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]

You need to follow these steps:

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\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]

  • Subtract row 1 from row 2 \left(R_2=R_2-R_1\right)

\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]

  • Subtract row 1 multiplied by 5 from row 3 \left(R_3=R_3-\left(5\right)R_1\right)

\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]

  • Subtract row 2 multiplied by 3 from row 1 \left(R_1=R_1-\left(3\right)R_2\right)

\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]

  • Subtract row 2 multiplied by 3 from row 3 \left(R_3=R_3-\left(3\right)R_2\right)

\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]

  • Multiply row 2 by 2 \left(R_2=\left(2\right)R_2\right)

\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]

  • Divide row 3 by −19 \left(R_3=\frac{R_3}{-19}\right)

\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]

  • Subtract row 3 multiplied by 16 from row 1 \left(R_1=R_1-\left(16\right)R_3\right)

\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]

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\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]

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7 0
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