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2 years ago

15

What is the sum of the solutions of the 2 equations? 4x = 12 & 2y + 10 = 22

1 answer:

KengaRu [80]2 years ago

4 0

Answer:

x = 3 & y = 6

Step-by-step explanation:

4x = 12

12/4 = 3

x = 3

2y + 10 = 22

22 - 10 = 12

12/2 = 6

y = 6

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The length of a rectangle is 2 ft shorter than 5 times its width, x. The area of the rectangle is less than 100 f12.

Ierofanga [76]

Answer:

A

Step-by-step explanation:

Area of rectangle = length x width

length = 5x - 2

width = x

area = (5x - 2) × x

100 = 5x² - 2x

the width would be less than the area of the rectangle, thus the inequality sign to be used is <

5x² - 2x < 100

7 0

3 years ago

Please answer only number seven.<br> Also, tell me the final answer and the process please

Sergeeva-Olga [200]

Left side is 22.
22 x 10 = 220
(20 - 6) x 12 = 168
6 x 20 = 120
220 + 168 + 120 = 508
Thank you very much.

7 0

3 years ago

Mick paid 2.94 in sales tax on an item that cost $42.00 before tax.at that rate how much would he pay in sales tax for an item t

Soloha48 [4]

0ne dollars and forty- seven cents

4 0

3 years ago

Combine like radicals to get your answer. <br><br> image attached

spin [16.1K]

Answer:

So the final answer is

$-8\sqrt{3}$

Step-by-step explanation:

Radicals:

In mathematics, a radical expression is defined as any expression containing a radical (√) symbol. Many people mistakenly call this a 'square root' symbol, and many times it is used to determine the square root of a number. However, it can also be used to describe a cube root, a fourth root, or higher.

The given expression is

$-5\sqrt{3} -3\sqrt{3}$

Now take common radical out so we will get

$\sqrt{3}(-5-3)$

Now add the Parenthesis part.

$\sqrt{3}(-8)$

So the final answer is

$-8\sqrt{3}$

5 0

3 years ago

Perform the indicated row operations, then write the new matrix.

Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

$\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]$

Operations:

$2R_1 + R_2 ->R_2$

$-3R_1 +R_3 ->R_3$

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

Step-by-step explanation:

The first operation:

$2R_1 + R_2 ->R_2$

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

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

Multiply by 2

$2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]$

$\left[\begin{array}{ccc}0&5&7|1\end{array}\right]$

The second operation:

$-3R_1 +R_3 ->R_3$

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

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

Multiply by -3

$-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]$

Add to row 2 elements are: $\left[\begin{array}{ccc}3&2&4|1\end{array}\right]$

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

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

Hence, the new matrix is:

$\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]$

3 0

3 years ago

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