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Umnica [9.8K]
3 years ago
8

(7x −4 y 15 )(4x 14 y −3 )

Mathematics
1 answer:
LenKa [72]3 years ago
8 0

Answer:

=11x+10y+12

Step-by-step explanation:

7x +4x -4y+14y 15-3

11x 10y 12

=11x+10y+12

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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.
sveticcg [70]

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]

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

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,

\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]·\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a=\frac{2}{11}; b=\frac{7}{33}; c=\frac{-1}{11} and d=\frac{-3}{11}. Substituting:

X= \left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11}  \end{array}\right]·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Multiplying the matrices, we have

X=\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11}  \end{array}\right]

6 0
3 years ago
Pls do it plz plz plz DO IT THE NOW!!!!
Zigmanuir [339]

Answer:

(2 , 30)   & (4, 60)

Step-by-step explanation:

y = 15x

When x = 2 , y = 15*2 = 30

When x = 4 ;   y = 15 *4 = 60

8 0
3 years ago
HELP THESE MATH QUESTIONS PLEASEEEEEEEE!!!<br> ASAPPPPP!!
PilotLPTM [1.2K]
The answer is y = +8

Explanation:
Since they tell us what x equals (3), let's plug that into the equation

y = (3) + 5

Now, we can just add 3 and 5 together, which we get 8

y = +8
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3 years ago
For a class project, a teacher cuts out 15 congruent
maw [93]

Answer:

its 60-15(pie) sq inches

Step-by-step explanation:

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Write two decimals one repeating and one terminating with the values between 0 and 1
weqwewe [10]

A repeating decimal is one that essentially goes on forever. A terminating decimal is one that has an end, therefore a definite value.

The fraction 1/3 is a repeating decimal, because when you divide 1 by 3, you get .333333 (to infinity). To show that something is repeating, draw a bar (or line) above the number that is repeating, in this case, 3.

The fraction 1/4 is a terminating decimal. Like the one above, when you divide 1 by 4, you get a fraction. In this case, it is .25, which does not repeat.

The fractions are there just to show you how you could get to either, but your terminating decimal is .25, and your repeating decimal is .3 (but with a line over the 3 if possible).

3 0
2 years ago
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