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kifflom [539]
3 years ago
14

What is the product of the two solutions from the quadratic formula?

Mathematics
1 answer:
ollegr [7]3 years ago
5 0

Answer:

c/a

Step-by-step explanation:

The product of two solutions from the quadratic formula is c/a.

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The solutions to the equation 2x2+x-1=2 are X= -3/2 or x=
Brilliant_brown [7]

Answer:

positive 3/2

Step-by-step explanation:

3 0
3 years ago
Which of the following is a counterexample to the statement “All even numbers are divisible by 4”? 2 x 2 = 4 7 is not divisible
Naily [24]
10 is not divisible by 4
5 0
3 years ago
Rewrite 81 as 3 to some power.
NikAS [45]
First, let's find all the factors of 81:

81/3=27
27/3=9
9/3=3
3/1=1

So the factors of 81 are 3,3,3,3, (the numbers by which we divided here).

This means we can write down 81 as 3*3*3*3 or 3^{4} - which is the answer to the question! (three to the power of 4)
5 0
3 years ago
[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation
irina1246 [14]

Answer:

y=-\frac{158}{579}

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]

And the vector B is formed with the solution of each equation of the system:b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called A_{2}.

A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]

The value of y using Cramer's rule is:

y=\frac{det(A_{2}) }{det(A)}

Find the value of the determinant of each matrix, and divide:

y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}

y=-\frac{158}{579}

7 0
3 years ago
What is 2/9 divided by 3?
Kryger [21]
((2÷9))÷3
(0.222)÷3
=0.074
3 0
4 years ago
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