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Mademuasel [1]
3 years ago
9

Determine whether each argument is valid. if the argument is valid, give a proof using the laws of logic. if the argument is inv

alid, give values for the predicates p and q over the domain {a, b} that demonstrate the argument is invalid. (a) ∃x (p(x) ∧ q(x)) ∴ ∃x q(x) ∧ ∃x p(x) hegg
Mathematics
1 answer:
Ede4ka [16]3 years ago
6 0
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Algebr<br>What is the product of the binomials?<br>(3x + 8)(5x - 9)​
vichka [17]

Answer:

x=-1

Step-by-step explanation:

7 0
3 years ago
Use matrices to solve the system of equations if possible. Use Gaussian elimination with back substitution or gauss Jordan elimi
CaHeK987 [17]

In matrix form, the system is given by

\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}

I'll use G-J elimination. Consider the augmented matrix

\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]

• Multiply through row 1 by -1.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]

• Multiply row 3 by 1/3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]

Then the solution to the system is

\boxed{x=9, y=-3, z=8}

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]

The third row tells us that z=8. Then in the second row,

y-z = -11 \implies y=-11 + 8 = -3

and in the first row,

x-y+z=20 \implies x=20 + (-3) - 8 = 9

5 0
2 years ago
Can someone help with this problem?
quester [9]

Answer:

0.3

Step-by-step explanation:

4 0
3 years ago
Please answer (middle school) (probability)
Dafna11 [192]
12 diffrent variations
8 0
2 years ago
Read 2 more answers
Make y the subject<br> y-aw=2w-1
Fudgin [204]

Answer:

y = 2w - 1 + aw

Step-by-step explanation:

Given

y - aw = 2w - 1 ( isolate y by adding aw to both sides )

y = aw - 1 + aw

8 0
3 years ago
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