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4 years ago

8y - 1-/2 ( 4y - 16)

Mathematics

1 answer:

Dimas [21]4 years ago

4 0

Answer:

2.(3y+4)

Step-by-step explanation:

1/2

8y-1/2.(4y -16))

3.1

4y-16=4.(y-4)

5.1 6y +8 =2•(3y+4)

final answer 2•(3y+4)

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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· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =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= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

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3 years ago

What will happen to the measures of central tendency if the outlier is removed

frosja888 [35]

It decreases the number of data by one.

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2 years ago

The radius of the base of a cylinder is decreasing at a rate of 121212 kilometers per second. The height of the cylinder is fixe

WARRIOR [948]

Answer:

7536 $km^3/sec$

Step-by-step explanation:

Given that:

Rate of decreasing of radius = 12 km/sec

Height of cylinder is fixed at = 2.5 km

Radius of cylinder = 40 km

To find:

The rate of change of Volume of the cylinder?

Solution:

First of all, let us have a look at the formula for volume of a cylinder.

$Volume = \pi r^2h$

Where $r$ is the radius and

$h$ is the height of cylinder.

As per question statement:

$r$ = 40 km (variable)

$h$ = 2.5 (constant)

$\dfrac{dV}{dt} = \dfrac{d}{dt}\pi r^2h$

As $\pi, h$ are constant:

$\dfrac{dV}{dt} = \pi h\dfrac{d}{dt} r^2\\\Rightarrow \dfrac{dV}{dt} = \pi h\times 2 r\dfrac{dr}{dt} \\$Putting the values:$\\\Rigghtarrow\dfrac{dV}{dt} = 3.14 \times 2.5\times 2 \times 40\times 12 \\\Rigghtarrow\dfrac{dV}{dt} = 7536\ km^3/sec$