Jason solved the following equation to find the value for x.

Mathematics

1 answer:

Ainat [17]1 year ago

5 0

Plug in 6.5 for x.
-8.5x-3.5x=-78
-8.5(6.5)-3.5(6.5)=-78
-55.25-22.75=-78
-78=-78

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Simplify the expression

finlep [7]

Answer:

$\frac{2*x - 2}{2*x} - \frac{3*x + 2}{4*x} = \frac{x - 6}{4*x}$

Step-by-step explanation:

We have the expression:

$\frac{2*x - 2}{2*x} - \frac{3*x + 2}{4*x}$

The first thing we want to do, is to have the same denominator in both equations, then we need to multiply the first term by (2/2), so the denominator becomes 4*x

We will get:

$(\frac{2}{2} )\frac{2*x - 2}{2*x} - \frac{3*x + 2}{4*x} = \frac{4*x - 4}{4*x} - \frac{3*x + 2}{4*x}$

Now we can directly add the terms to get:

$\frac{4*x - 4}{4*x} - \frac{3*x + 2}{4*x} = \frac{4*x - 4 - 3*x - 2}{4*x} = \frac{x - 6}{4*x}$

We can't simplify this anymore

3 0

3 years ago

For this exercise assume that all matrices are ntimesn. Each part of this exercise is an implication of the form "If "statement

inna [77]

Answer:

C. True; by the Invertible Matrix Theorem if the equation Ax=0 has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the n x n identity matrix.

Step-by-step explanation:

The Invertible matrix Theorem is a Theorem which gives a list of equivalent conditions for an n X n matrix to have an inverse. For the sake of this question, we would look at only the conditions needed to answer the question.

There is an n×n matrix C such that CA= $I_n$ .
There is an n×n matrix D such that AD= $I_n$ .
The equation Ax=0 has only the trivial solution x=0.
A is row-equivalent to the n×n identity matrix $I_n$ .
For each column vector b in $R^n$ , the equation Ax=b has a unique solution.
The columns of A span $R^n$ .

Therefore the statement:

If there is an n X n matrix D such that AD=I, then there is also an n X n matrix C such that CA = I is true by the conditions for invertibility of matrix: