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LenKa [72]
3 years ago
6

Find the domain and range of the INVERSE of the given function. Square root x-3

Mathematics
1 answer:
Ulleksa [173]3 years ago
4 0
f(x)=\sqrt{x-3}\\
y=\sqrt{x-3}\\
y^2=x-3\\
x=y^2+3\\
f^{-1}(x)=x^2+3\\
D:x\in\matxbb{R}\\
R:y\in\langle3,\infty)
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What is the solution to the system of linear equations? y=4x and y-4x=0​
musickatia [10]

Answer:

  y = 4x; there are an infinite number of solutions

Step-by-step explanation:

Both equations describe the same relation between x and y. The system is "dependent", so has an infinite number of solutions. For any value of x, the solution is ...

  (x, y) = (x, 4x)

7 0
3 years ago
I need help with this one question<br> 8r - 5q = 3<br> solve for q <br> how do i go about this
KATRIN_1 [288]
8r - 3 = 5q --> q = (8r - 3)/5
6 0
3 years ago
How do I do this because it's really confusing haha
netineya [11]
A = 1.5
B = -1/3
C = -4/3

A is a positive so there is only one possibility
B is in between 0 and -1, so it has to be -1/3
C is below -1 and so its the improper fraction

5 0
3 years ago
There are 54 students in a class. The Venn diagram below shows how many students play a sport, take a foreign language, do both,
andrew11 [14]

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                                         |         |

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4 0
2 years ago
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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