Ask question

Ask question

All categories

2 years ago

15

I need help with this question

1 answer:

zhuklara [117]2 years ago

7 0

Answer: 5x - 1 = -6

Step-by-Step Explanation:

We have this diagram here showing numbers and symbols. Just a remainder, "X" is a variable which can be found through and equation of Algebra. Alright, so we have 5 boxes with the symbol "X". Alright, that's a start, so we have <u>5x</u> as a start of the equation. We have <u>-1</u> on the same side as well.

Alright, so on the other side- we have Six symbols of "-1".

That concludes to... <u>-6</u>. Alright, notice how all of the symbols are on a Scale, so we can heavily imply that the "=" symbol will take place between the two sets of symbols. Alright, combining them we get.. <u>5x - 1 = -6.</u>

<em>If you need further instructions/help, don't hesitate to give me a shout out. Bye, and I hope this helps.</em>

You might be interested in

Help solve algebra 15a=6a-90

AleksandrR [38]

15a =6a - 90

Move all the numbers with variables to one side

15a - (6a) = 6a -(6a) -90
9a = -90
9a/9 = -90/9
a = -9

Hope this helps

5 0

3 years ago

7 is a factor of which number?<br> 28<br> 16<br> 11<br> 4 4

maw [93]

Answer:

28

Step-by-step explanation:

28 = 7 x 4

So,

7 is a factor of 28.

6 0

2 years ago

A number is equal to the sum of half a second number and 3. The first number is also equal to the sum of one-quarter of the seco

IrinaK [193]

First number=1/2x+3
first number=1/4x+5

7 0

3 years ago

Please answer all of these

FromTheMoon [43]

What’s the questions

7 0

2 years ago

Read 2 more answers

Arrange the entries of matrix A in increasing order of their cofactors values

givi [52]

To find the cofactor of

$A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]$

We cross out the Row and columns of the respective entries and find the determinant of the remaining $2\times 2$ matrix with the alternating signs.

$Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|$

$Ac_{11}=4\times 1- -1\times 2$

$Ac_{11}=4+ 2$

$Ac_{11}=6$

$Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|$

$Ac_{12}=-(-7\times 1- -1\times -8)$

$Ac_{12}=-(-7- 8)$

$Ac_{12}=15$

$Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|$

$Ac_{21}=-(5\times 1- 3\times 2)$

$Ac_{21}=-(5-6)$

$Ac_{21}=1$

$A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|$

$Ac_{23}=-(7\times 2 -8\times 5)$

$Ac_{23}=-(14-40)$

$Ac_{23}=26$

$A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|$

$Ac_{31}=5\times -1 -4\times 3$

$Ac_{31}=-5-12$

$Ac_{31}=-17$

$A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|$

$Ac_{33}=7\times 4- -7\times 5$

$Ac_{33}=28+35$

$Ac_{33}=63$

Therefore in increasing order, we have;

$Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63$

7 0

3 years ago

Read 2 more answers