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aalyn [17]
3 years ago
14

Let f(x) = x - 2 and g(x) = x2 - 7x - 9. Find f(g(-1))

Mathematics
1 answer:
Montano1993 [528]3 years ago
4 0
f(x)=x-2 \\
g(x)=x^2-7x-9 \\ \\
g(-1)=(-1)^2-7 \times (-1)-9 = 1+7-9=-1 \\
f(g(-1))=f(-1)=-1-2=-3 \\ \\
\boxed{f(g(-1))=-3}
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Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
Alona [7]

Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

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3 years ago
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BlackZzzverrR [31]

Answer:

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Step-by-step explanation:

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3 years ago
Answer asap 50 point's brainlyest.
Anni [7]

Answer:

Option B

The answer is 16

Step-by-step explanation:

<h3><u>Given</u>;</h3>
  • 4(v – 13) = 12
<h3><u>To </u><u>Find</u>;</h3>
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Now,

4(v – 13) = 12

4(v – 13)/4 = 12/4

v – 13 = 3

v – 13 + 13 = 3 + 13

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Thus, The value of v is 16

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What is the mode for the month?
Papessa [141]

Answer:the answer would be 20

Step-by-step explanation:

because it is use the most

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Charra [1.4K]

Answer:

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