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2 years ago

Find the result in the simplest form: 2 - (1 - 2i) + (4 - 5i) - (1 - 3i)

Mathematics

1 answer:

never [62]2 years ago

4 0

Answer:

Step-by-step explanation:

Given

2 - (1 - 2i) + (4 - 5i) - (1 - 3i) ← distribute parenthesis

= 2 - 1 + 2i + 4 - 5i - 1 + 3i ← collect like terms

= 4

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A circle has a radius of 10 inches. Find the approximate length of the arc intersected by a central angle of 2π/3

Aleksandr-060686 [28]

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Radius of circle = 10 inches

Central angle is given by

$\frac{2\pi }{3}$

As we know the formula for " Length of an arc " :

Length of arc is given by

$l=r\theta\\\\l=10\times \frac{2\pi}{3}\\\\l=10\times \frac{2}{3}\times \frac{22}{7}\\\\l=20.95\ inches$

Hence, length of an arc is 20.95 inches.

Hence, Option 'C' is correct.

5 0

2 years ago

Read 2 more answers

Help. Urgent. Skenekeks

9966 [12]

Answer:

Step-by-step explanation:

Plug in the numbers for the x-value and solve the equation.

| 3(80/3)/4 + 1 | = 16

| 3(-40/3)/4 + 1 | = 16

6 0

2 years ago

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$