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1 year ago

What is the Value of the underlined digit 646,950

Mathematics

2 answers:

Jobisdone [24]1 year ago

4 0

I'm sorry, but what digit is underlined?

san4es73 [151]1 year ago

3 0

You don't have any underlined digits but heirs all of them.
(646,950) the first 6 is in the one hundred thousands place
(46,950) the 4 is in the ten thousands place.
(6,950) the 6 is in the thousands place
(950) the 9 is in the hundreds place.
(50) the 5 is in the tenth place.
(0) and the 0 well you should know that .

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Suppose the coefficient matrix of a system of linear equations has a pivot position in every row. Explain why the system is cons

kap26 [50]

Answer:

If in each row of the supposed coefficient matrix, there is a pivot position. Therefore, it is true that the bottom row of the coefficient matrix also has a pivot position. As a result, there will not be space for the augmented column to have a. Thus, we say the system is consistent.

Step-by-step explanation:

In the problem, we have a coefficient matrix comprising linear equations. If in each row of the supposed coefficient matrix, there is a pivot position. Therefore, it is true that the bottom row of the coefficient matrix also has a pivot position. As a result, there will not be space for the augmented column to have a. Thus, we say the system is consistent based on the theorem.

7 0

1 year ago

Given matrix A below, and that A = B, find the value of the elements in B. A = 9 −2 3 2 17 0 3 22 8 b11 = b12 = b13 = b21 = b22

GuDViN [60]

Answer:

$b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8$

Step-by-step explanation:

Consider the given matrix

$A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]$

Let matrix B is

$B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]$

It is given that

$A=B$

$\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]$

On comparing corresponding elements of both matrices, we get

$b_{11}=9,b_{12}=-2,b_{13}=3$

$b_{21}=2,b_{22}=17,b_{23}=0$

$b_{31}=3,b_{32}=22,b_{33}=8$

Therefore, the required values are $b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8$ .

3 0

2 years ago

Read 2 more answers

Joseph received a $20 gift card for downloading music. Each downloaded song costs $1.29. Explain how to write and solve an inequ

poizon [28]

Let x be the number of songs downloaded.
Each song is $1.29; the total cost would be found by multiplying the cost by the number of songs, or 1.29x.
This cannot be more than 20, so we set this less than or equal to 20:
1.29x ≤ 20.

To solve this, we divide both sides by 1.29: 
 $\frac{1.29x}{1.29}$ ≤ $\frac{20}{1.29}$ ;
x ≤15.5.

We cannot download half of a song, so we round this down to 15 (although the number rounds up mathematically, he would not have enough money to download 16 songs). This means he can download at most 15 songs.

8 0

2 years ago

Read 2 more answers

Answer below these questions

timurjin [86]

Answer:

A: 6x⁸y⁵

B: 4x⁵z⁸

C: 48a¹²b⁵

D: 6s⁹t³

Step-by-step explanation:

When you multiply 2 exponents together, you add them. When you power an exponent, you multiply the 2 exponents together,

3x²2y⁴(2x⁶y)

6x⁸y⁵

xz³(4x⁴z⁵)

4x⁵z⁸

(4a³)²(3a⁶b⁵)

16a⁶(3a⁶b⁵)

48a¹²b⁵

6s⁵t(s⁴t²)

6s⁹t³

3 0

2 years ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

3 0

1 year ago