1) x = 5y - 30
2) x = - 1.6y + 3.3
<em>im assuming these are 2 different questions</em>
<h3>
Explanation:</h3><h3>1)</h3>
Start with the equation:
x - 5y= -30
add 5y to both sides:
x = 5y - 30
Your answer is x = 5y - 30
<h3>2)</h3>
Start with the equation:
3x+5y=10
subtract 5y from both sides:
3x = -5y + 10
divide by 3 to get x singular
x = - 1.6y + 3.3
Your answer is x = - 1.6y + 3.3
Hope this helps :)
<u>ANSWER:</u>
The solution set for the inequality 7x < 7(x - 2) is null set 
<u>SOLUTION:</u>
Given, inequality expression is 7x < 7 × (x – 2)
We have to give the solution set for above inequality expression in the interval notation form.
Now, let us solve the inequality expression for x.
Then, 7x < 7 × (x – 2)
7x < 7 × x – 2 × 7
7x < 7x – 14
7x – (7x – 14) < 0
7x – 7x + 14 < 0
0 + 14 < 0
14 < 0
Which is false, so there exists no solution for x which can satisfy the given equation.
So, the interval solution for given inequality will be null set
Hence, the solution set is 
Answer:
Infinite number of solutions.
Step-by-step explanation:
We are given system of equations



Firs we find determinant of system of equations
Let a matrix A=
and B=![\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%5C%5C1%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)


Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.
We are finding rank of matrix
Apply
and 
:![\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C1%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
Apply
:![\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C6%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
Apply 
:![\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C6%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
Apply
and 
:![\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C%5Cfrac%7B13%7D%7B2%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Apply 
:![\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B9%7D%7B2%7D%5C%5C%5Cfrac%7B13%7D%7B2%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.
Therefore, rank of matrix is equal to rank of B.
Answer:
Step-by-step explanation:
3 Triangle
4 Quadrilateral
5 Pentagon
6 Hexagon
7 Heptagon
8 Octagon
9 Nonagon
10 Decagon
Answer:
−y−1=3
Step-by-step explanation:
Which equation can be used to find the solution of (1/4)^y+1=64?
This can be solved by power of indices
(1/4)^(y+1)=64
(4^-1)^(y + 1)= 4^3
Note
(x^a)^b = x^ab
Hence:
4^(-1)(y + 1)= 4^3
4^-y - 1 = 4^3
Divide both sides by 4
−y−1=3
Hence, the equation that can be used to find the solution of (1/4)^y+1=64 is
−y−1=3