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3 years ago

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:

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

canceled 2

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

Method 2:

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

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

use the distributive property a(b + c) = ab + ac

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

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· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

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

I = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

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)

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:

Step-by-step explanation:

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

Basically the absolute value of N1 - N2

3 0

3 years ago

Read 2 more answers

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​

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