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2 years ago

How do you explain -3 1/3 / 4/9

Mathematics

1 answer:

solniwko [45]2 years ago

8 0

Step-by-step explanation:

i have just explained everything

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A train has 6 passenger cars and each car has 4 columns of seats and each column holds 50 passengers. If one of the cars has 25

Tema [17]

Given:
The train has 6 passenger cars
Each car has 4 columns
And 1 column hold 50 passengers

So, total of seats will be = 6 * 4 * 50 = 1200

Now, there are 25 empty seats,

therefore the correct equation will be :

Number of seats per car: 4 × 50 = 200

Total number of seats: 200 × 6 = 1,200

Number of passengers: 1,200 − 25 = p

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2 years ago

2)<br> 3<br> 3 cm<br> 9 cm<br> Find area

sergey [27]

Answer:
27cm^2

Explanation:
3 x 9 = 27

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2 years ago

Help maths i’m bad at this :)))(

Tju [1.3M]

I suck at math to I’m so so sorry Ask a tutor they will help

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2 years ago

Sin4u=2sin 2u cos 2u

yKpoI14uk [10]

Answer:

If you've learnt sin(A+B) = sinAcosB + cosAsinB,

sin(4u)

= sin(2u+2u)

= sin(2u)cos(2u) + cos(2u)sin(2u)

= 2 sin(2u) cos(2u).

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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

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

As can be seen, we have obtained the identity matrix to the left. So, we are done.

6 0

2 years ago