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2 years ago

What are the dimensions of the matrix?

Mathematics

2 answers:

mrs_skeptik [129]2 years ago

6 0

For this case we have that by definition, the dimension of a matrix is given by:

<em>The number of rows by the number of columns </em>

It is observed that we have a column that consists of three rows.

$\left[\begin{array}{c}6&\\2&\\8&\end{array}\right]$

Thus, the dimension of the matrix is 3 * 1

Answer:

3 * 1

Option A

denis-greek [22]2 years ago

3 0

Answer: 3 x 1

There are 3 rows and 1 column. We write the number of rows first. The rows are laid out horizontally while the columns are vertical. We say this is a "3 by 1 matrix".

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Which expression is equivalent to 2/3(6x+3)

irina1246 [14]

2/3(6x+3)
2/3 x 3(2x+1)
2(2x+1)
4x+2
Therefore your answer is 4x+2

5 0

2 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

2 years ago

A collection of five coins has a value of $1 and consists of dimes and

DENIUS [597]

For this case we propose a system of equations. We have to:

x: Let the variable that represents the number of dimes

y: Let the variable that represents the number of quaters

We know that:

One dime equals 10 cents, $0.10

A quater equals 0.25 cents, $0.25

According to the statement we have:

$x + y = 5\\0.10x + 0.25y = 1$

We multiply the first equation by -0.10:

$-0.10x-0.10y = -0.5$

We have the following equivalent system:

$-0.10x-0.10y = -0.5\\0.10x + 0.25y = 1$

We add the equations:

$-0.10x + 0.10x-010y + 0.25y = -0.5 + 1\\0.15y = 0.5\\y = \frac {0.5}{0.15}\\y = 3.33$

Approximately 3 quater coins

$x = 5-y = 5-3 = 2$

And two dimes

Answer:

3 quater

2 dimes

4 0

2 years ago

Help plzzzz I need help ASAP thank you

pentagon [3]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What is the horizontal distance between the two points (-7,-4) and (12,-4)? Remember that distance is always positive! Hint: The

RSB [31]

<h3>Answer: 19</h3>

==================================================

Explanation:

Draw out a number line. Plot -7 and 12 on the number line. Draw in the tickmarks between them. You should find the distance from -7 to 12 is 19 units since you need to go 19 spaces from either -7 to 12, or vice versa.

You can use subtraction to get

-7-12 = -19

12-(-7) = 12+7 = 19

The final result is made positive since negative distance does not make sense.

So you'd have |-7-12| = |-19| = 19 or |12-(-7)| = |19| = 19.

All of this only works because the two y coordinates are the same, which makes a horizontal line through the two given points.

7 0

3 years ago