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3 years ago

7

99 POINTS.... PLEASEEEEE HELP !!!

1 answer:

Nookie1986 [14]3 years ago

5 0

You have a lot of questions here, I can help with some.

1) To find the inverse, switch the x and f(x) and solve for f(x).
x = 3f(x) - 7
x + 7 = 3(f(x)
(x + 7) / 3 = f(x)

2) It is a function, because each input pairs with only 1 output.

3) Any diagonal line would be a function.

4 - 8 (We can't do without the diagram)

9) f(-9) = 3(-9) - 6
f(-9) = -27 - 6
f(-9) = -33

10) f(-2) = 42 + 3
f(-2) = 45

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2 years ago

Solving for matrices

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Step-by-step explanation:

The augmented matrix for the system of three equaitons is

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\5&2&-2&|&11\\5&-4&4&|&-7\end{array}\right)$

Multiply the first row by 5, the second row by -3 and add these two rows:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\5&-4&4&|&-7\end{array}\right)$

Subtract the third row from the second:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&6&-6&|&18\end{array}\right)$

Divide the third row by 6:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&1&-1&|&3\end{array}\right)$

Now multiply the third equation by 26 and add it to the second row:

$\left(\begin{array}{ccccc} 3&-4&-5&|&-27\\0&-26&-19&|&-168\\0&0&-45&|&-90\end{array}\right)$

You get the system of three equations:

$\left\{\begin{array}{r}3x-4y-5z=-27\\-26y-19z=-168\\-45z=-90\end{array}\right.$

From the third equation

$z=\dfrac{90}{45}=2.$

Substitute z=2 into the second equation:

$-26y-19\cdot 2=-168\\ \\-26y-38=-168\\ \\-26y=-168+38=-130\\ \\y=\dfrac{130}{26}=5.$

Now substitute z=2 and y=5 into the first equation:

$3x-4\cdot 5-5\cdot 2=-27\\ \\3x-20-10=-27\\ \\3x-30=-27\\ \\3x=-27+30=3\\ \\x=1.$

The solution is (1,5,2)

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3 years ago