All categories

3 years ago

Subtract. (3x+4)-(x+2)

1 answer:

8 0

I hope this helps please tell me if I got it wrong so I can fix and learn more lol

I would first distribute the negative to x+2 so then it becomes x-2.
The equation that I did was 3x+4-x-2 And then u would subtract 3x-x which would become 2x and then just subtract 4-2 And that would become 2

Answer: 2x+2

You might be interested in

2. Four less than six times a number is 32. Write and solve an equation.

zhenek [66]

Answer:

6x - 4 = 32. The variable (times a number) is equal to 6.

Step-by-step explanation:

So this equation is basically asking you to formulate an equation using only the words given. For this equation, let's use a variable in place of the part where it says "times a number." When a question says that phrase, that means that they are telling you to use a variable in place of a mystery number. A variable is a letter in place of a mystery number. Knowing this, let's now solve the equation. We will use the variable x in place of our mystery number.

Four less than 6 times a number: 6x - 4 _ __.

Now we can add the equal sign and the 32: 6x - 4 = 32.

We can now solve. Add 4 to both sides of the equation to cancel out the -4:

6x - 4 = 32

6x = 36

Now we divide 6 into both sides of the equation to single out the x.

6 6

36/6 = 6, so:

x = 6

I hope that this helps.

8 0

2 years ago

An electronics company produces transistors, resistors, and computer chips. Each transistor requires 3 units of copper, 2 unit

Gala2k [10]

Answer:

475 transistors, 25 resistors and 50 computer chips can be produced.

Step-by-step explanation:

Let us consider, p = Number of transistors.

q = Number of resistors.

r = Number of computer chips.

The following three linear equations according to question,

$3\times p + 3\times q + 2\times r = 1600\\2\times p + 1\times q + 1\times r = 1025\\1\times p + 2\times q + 2\times r = 625$

The matrix form of any system, Ax = B

Where, A = Coefficient matrix

B = Constant vector

x = Variable vector

$A = \left[\begin{array}{ccc}3&3&2\\2&1&1\\1&2&2\end{array}\right], x = \left[\begin{array}{ccc}p\\q\\r\end{array}\right], B = \left[\begin{array}{ccc}1600\\1025\\625\end{array}\right]$

The inverse matrix, $A^{-1}$ can be found by using the following formula,

$A^{-1} = \frac{1}{det A}\times (C_{A}) ^{T}$

Where, det A = Determinant of matrix A.

$C_{A}$ = Matrix of cofactors of A

Now, applying this formula to find $A^{-1}$ ;

$det A = \left[\begin{array}{ccc}3&3&2\\2&1&1\\1&2&2\end{array}\right] = 3\times(2-2)-3\times(4-1)+2\times(4-1) = -3$

Here, $det A\neq 0$ , thus the matrix is invertible.

$C_{A} = \left[\begin{array}{ccc}(2-2)&-(4-1)&(4-1)\\-(6-4)&(6-2)&-(6-3)\\(3-2)&-(3-4)&(3-6)\end{array}\right] = \left[\begin{array}{ccc}0&-3&3\\-2&4&-3\\1&1&-3\end{array}\right] \\(C_{A}) ^{T} = \left[\begin{array}{ccc}0&-3&3\\-2&4&-3\\1&1&-3\end{array}\right] ^{T} = \left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right]$

$A^{-1} = \frac{1}{-3}\times\left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right] \\ So, x= \frac{1}{-3} \left[\begin{array}{ccc}0&-2&1\\-3&4&1\\3&-3&-3\end{array}\right]\times\left[\begin{array}{ccc}1600\\1025\\625\end{array}\right]= \frac{1}{-3} \left[\begin{array}{ccc}-1425\\-75\\-150\end{array}\right] = \left[\begin{array}{ccc}475\\25\\50\end{array}\right]$