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ddd [48]
2 years ago
15

I need help with this question

Mathematics
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>

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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
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