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Semmy [17]
3 years ago
12

How do you solve -8 + 5x ≥ -13.

Mathematics
2 answers:
Studentka2010 [4]3 years ago
8 0

Answer:  x ≥ -1

Step-by-step explanation:

-8 + 5x ≥ -13

<em>First, we plus 8 on both sides</em>

5x ≥ -5

<em>And then we divide 5 from both sides</em>

x ≥ -1

Evgen [1.6K]3 years ago
7 0

Answer:

x≥-1

Step-by-step explanation:

5x≥-5

x≥-1

Add like terms which are -8 and -13

that gives you -5

Divide 5 on both sides to isolate x and you have x is greater than or equal to -1

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Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.
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The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =<em>I</em>.

First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

So, \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Isolating the X, we have

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right] -  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Resolving:

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]

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Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X=\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]⁻¹·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

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Multiplying the matrices, we have

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