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Alja [10]
3 years ago
14

Which two expressions are equivalent? B. A 4(2 + x) 4.2 +2.x 4 + 2 + x (4 + 2) + X D. C. 4. X + 2 4. (x + 2) 4 = (2-x) 4- 2 = X​

Mathematics
1 answer:
Nesterboy [21]3 years ago
4 0
I think it’s B because it’s in the right for mate
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For free points, Please solve, -5×3÷(2-(-4)-3×(-3))? ​
Aleksandr [31]

Answer:

\Huge \boxed{-1}

Step-by-step explanation:

\displaystyle \frac{-5 \times 3}{2-(-4) - 3 \times (-3)}

Solving for brackets and multiplying.

\displaystyle \frac{-15}{2+4 +9}

Adding the numbers in the denominator.

\displaystyle \frac{-15}{15}

Dividing.

=-1

3 0
3 years ago
Read 2 more answers
The product of 1 1/2 and 2 is
Over [174]

Method 1:

Convert the mixed number to the improper fraction:

1\dfrac{1}{2}=\dfrac{1\cdot2+1}{2}=\dfrac{3}{2}

Make the product:

1\dfrac{1}{2}\cdot2=\dfrac{3}{2}\cdot2

<em>canceled 2</em>

=\dfrac{3}{\not2_1}\cdot\not2^1}=\boxed{3}

Method 2:

1\dfrac{1}{2}=1+\dfrac{1}{2}

1\dfrac{1}{2}\cdot2=\left(1+\dfrac{1}{2}\right)\cdot2

<em>use the distributive property a(b + c) = ab + ac</em>

(1)(2)+\left(\dfrac{1}{2}\right)(2)=2+1=\boxed{3}

7 0
3 years ago
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
Let v⃗ 1=⎡⎣⎢033⎤⎦⎥,v⃗ 2=⎡⎣⎢1−10⎤⎦⎥,v⃗ 3=⎡⎣⎢30−3⎤⎦⎥ be eigenvectors of the matrix A which correspond to the eigenvalues λ1=−1, λ2
kaheart [24]

Answer:

- x as a linear combination :

x = -1 v1+ 0 v2+ 1 v3.

- Transpose Ax = (12, -6, -6)

Step-by-step explanation:

Given v1 = (0 3 3),v2 = (1 −1 0), v3 = (3 0 −3) be eigenvectors of the matrix A which correspond to the eigenvalues λ1 = −1, λ2 = 0, and λ3 = 1, respectively, and let x = (−2 −4 0). Express x as a linear combination of v1, v2, and v3, and find Ax .

To write x as a linear combination of v1, v2, and v3

x = -1 v1+ 0 v2+ 1 v3.

To find Ax

Write A = (0 ......3 ......3 )

...................(1 ......-1 ......0)

...................(3 ......0......-3)

Since transpose x = (-2, 4, 0)

Ax =......... (0 ......3......3 )(-2)

...................(1 ......-1 ......0)(4)

...................(3 ......0......-3)(0)

= (0×-2 + 3×4 + 3×0)

...(1×-2 + -1×4 + 0×0)

.. (3×-2 + 0×4 + -3×0)

As = (12)

....(-6)

....(-6)

Transpose Ax = (12, -6, -6)

7 0
3 years ago
Use the absolute value formula to express the distance between 9 and -8
Sonja [21]

Answer:

17

Step-by-step explanation:

| 9- (- 8) | = | 17 | = 17

Basically the absolute value of N1 - N2

3 0
3 years ago
Read 2 more answers
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