$-1(x+5)=3[x+2x-1)]\\\\\text{for}\ -1(x+5):\ \text{distribtutive property}\\\text{for}\ [x+(2x-1)]:\ \text{associative property}\\\\(-1)(x)+(-1)(5)=3[(x+2x)-1]\\-x-5=3(3x-1)\\\\\text{for}\ 3(3x-1):\ \text{distributive property}\\\\-x-5=(3)(3x)+(3)(-1)\\-x-5=9x-3\\\\\text{for the equation}:\ \text{addition property of equality}\\\\-x-5=9x-3\qquad\text{add 5 to both sides}\\-x-5+5=9x-3+5\\-x=9x+2\\\\\text{for the equation:}\ \text{subtraction property of equality}\\\\-x=9x+2\qquad\text{subtract}\ 9x\ \text{from both sides}$

$-x-9x=9x-9x+2\\-10x=2\\\\\text{for the equation:}\ \text{division property of equality}\\\\-10x=2\qquad\text{divide both sides by -10}\\\dfrac{-10x}{-10}=\dfrac{2}{-10}\\x=-0.2$

$\text{Distributive property:}\ a(b+c)=ab+ac\\\text{Associative property:}\ (a+b)+c=a+(b+c)$

6 0

2 years ago

One hen lays 7 eggs in three days.<br>How many eggs do three hens lay in nine days?

kramer

Answer:

63 eggs.

Step-by-step explanation:

7 times 3 equals 21, which is the amount of eggs one chicken lays in 9 days. Multiply that by 3, you get 63.

3 0

2 years ago

Read 2 more answers

For waht values of x do the vectors -1,0,-1), (2,1,2), (1,1, x) form a basis for R3?

DaniilM [7]

<h2>Answer:</h2>

The values of x for which the given vectors are basis for R³ is:

$x\neq 1$

<h2>Step-by-step explanation:</h2>

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

$\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]$

Now, the determinant of this matrix is:

$\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1$

Hence,

$-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1$

4 0

2 years ago