All Answers 358150 Answers

|2 -1 2|

|-4 0 1|

|-1 -1 -1|

(the vertical line at the front and back of each one are all suppose to be connected vertically)

$$\small{\text{
$
\left|
\begin{array}{rrr}
2 &-1 &2 \\
-4 &0 &1 \\
-1 &-1 &-1 \\
\end{array}
\right|
$
}}\\\\\\
\begin{array}{ll}
= &
2\cdot 0 \cdot (-1) + (-1)\cdot(-1)\cdot 1 + (-4)\cdot(-1)\cdot 2\\
&- (-1)\cdot 0 \cdot 2 - 2\cdot(-1)\cdot 1 - (-4)\cdot(-1)\cdot(-1)\\
= &
0 + 1 + 8\\
&-0 + 2 +4\\
= &
9\\
&+6\\
= 15
\end{array}
$
}}$$

y=3x+1 ( line )

y=x^2+1 ( parabola )

$$\small{\text{
$
\begin{array}{rcl}
y_{parabola} &=& y_{line} \\
x^2+1 &=& 3x+1\\
x^2+\not{1} &=& 3x+\not{1}\\
x^2-3x &=& 3x \\
x^2-3x &=& 0 \\
\underbrace{x}_{x_1=0}\cdot\underbrace{(x-3)}_{=0} &=& 0 \\
x_2-3 &=& 0 \\
x_2 &=& 3
\end{array}
$
}}$$

$$\small{\text{
$
\begin{array}{rcl}
\\y_1 &=& 3\cdot x_1 + 1 \\
y_1 &=& 3 \cdot 0 + 1 \\
y_1 &=& 1\\\\
y_2 &=& 3\cdot x_2 + 1 \\
y_2 &=& 3 \cdot 3 + 1 \\
y_2 &=& 10
\end{array}
$
}}$$

So we have: $$\small{\text{$(\ x_1 = 0 \quad y_1 = 1 \ )$}}$$ and $$\small{\text{$(\ x_2 = 3 \quad y_2 = 10 \ )$}}$$

Cramer's Rule

$$\begin{array}{rcrcrcr}
1\cdot x &+& 4 \cdot y &-&3 \cdot z &=& -8 \\
3\cdot x &-& 1 \cdot y &+& 3 \cdot z &=& 12 \\
1\cdot x &+& 1 \cdot y &+& 6 \cdot z &=& 1 \\
\end{array}$$

$$\small{\text{
$
x=
\frac
{\left|\begin{array}{rrr}
-8 & 4 & -3 \\
12 & -1 & 3 \\
1 & 1 & 6
\end{array}\right|}
{\left|\begin{array}{rrr}
1 & 4 & -3 \\
3 & -1 & 3 \\
1 & 1 & 6
\end{array}\right|}
$
}} \quad
\small{\text{
$
y=
\frac
{\left|\begin{array}{rrr}
1 & -8 & -3 \\
3 & 12 & 3 \\
1 & 1 & 6
\end{array}\right|}
{\left|\begin{array}{rrr}
1 & 4 & -3 \\
3 & -1 & 3 \\
1 & 1 & 6
\end{array}\right|}
$
}} \quad
\small{\text{
$
z=
\frac
{\left|\begin{array}{rrr}
1 & 4 & -8 \\
3 & -1 & 12 \\
1 & 1 & 1
\end{array}\right|}
{\left|\begin{array}{rrr}
1 & 4 & -3 \\
3 & -1 & 3 \\
1 & 1 & 6
\end{array}\right|}
$
}}$$

$$\small{\text{
$
x=
\frac
{
(-8)\cdot(-1)\cdot6+1\cdot4\cdot3+12\cdot 1\cdot(-3)
-1\cdot(-1)\cdot(-3)-(-8)\cdot 1\cdot 3 - 12\cdot 4\cdot 6
}
{
1\cdot(-1)\cdot 6 + 1\cdot 4 \cdot 3 + 3\cdot 1 \cdot (-3)
-1\cdot(-1)\cdot(-3) - 1\cdot 1 \cdot 3- 3\cdot 4 \cdot 6
}
=\frac
{48+12-36-3+24-288}
{-6+12-9-3-3-72}
=\frac
{-243}
{-81}
=3
$
}}$$

$$\small{\text{
$
y=
\frac
{
1\cdot 12 \cdot 6 + 1\cdot(-8)\cdot3 + 3\cdot 1 \cdot (-3)
-1\cdot 12 \cdot (-3) - 1\cdot 1 \cdot 3 - 3 \cdot (-8) \cdot 6
}
{
-81
}
=\frac
{72-24-9+36-3+144}
{-81}
=\frac
{216}
{-81}
=-2\frac{2}{3}
$
}} \quad$$

$$\small{\text{
$
z=
\frac
{
1\cdot(-1)\cdot 1 +1\cdot 4 \cdot 12 + 3 \cdot 1 \cdot(-8)
-1\cdot(-1)\cdot (-8) -1\cdot 1 \cdot 12 - 3\cdot 4 \cdot 1
}
{
-81
}
=\frac
{-1+48-24-8-12 -12}
{-81}
=\frac
{-9}
{-81}
=\frac{1}{9}
$
}} \quad$$

In real life No, but in math I am not sure.

She planted 9 in each row

I didn't know how to do this so I gave myself a quick lesson from here

Watch th video and work through the video example

then put it together with your knowledge of how elimination method to solve simultaneous equations works.

$$\begin{pmatrix}
2 &1&0&|-4 \\
0 &-2&4&|+0 \\
3 &0&-2&|-11 \\
\end{pmatrix}$$

First you need to get rid of the 3 in the bottom left corner.

1)  multiply R1 by3  and R3 by 2     WRITE down the new matrix

2)  R1-R3  replaces R3                    WRITE down the new matrix  (the 3 has been replaced with a 0)

NOW there is a 3 in R3, C2  You have to get rid of it

3) mult R2 by 3 and mult R3 by 2    WRITE down the new matrix

4) R2+R3 replaces R3                    WRITE down the new matrix  (the 6 has been replaced with a 0)

NOW the diagonal must only contain ones

5) Multiply R1 by 1/6,      Multiply R2 by -1/6         and multiply R3 by 20

And you should be able to take it from there

answer

x=-3,    y=2,   and    z=1

sally only has 36 trees, so how can she plant 144 trees on each row? @sweatyperson666

dozen=12

12*3=36

$${\mathtt{36}}{\mathtt{\,\times\,}}{\mathtt{4}} = {\mathtt{144}}$$

Hopefully i answered the first question

-0.666666...7 or  -4/6

I'm not quite sure what you're asking, but I believe you mean to simplify it. You need to find like terms, which are parts of a question that are similar enough to add. 4 and 5 would be like terms, and you could add them to get 9. In a similar way, x2 and 5x are two similar terms. Try to visualize something like boxes or silverware in a droor to help you. Simply, 2x + 5x would equal 7x. Since we don't know what "x" equals, the simplest form of that equation is 7x+6.

You have to make the bottoms (denominators) the same.  It will be 24

Look at this

$$(x2-x1)^{2}+(y2-y1)^{2}$$, then square root the resulting answer

3x12 = 36

28.97/36 = 0.805

see a doctor

The calculator at this site says:  50 031 545 098 999 707

The area of the base is 4 x 4 = 16 square inches.

Also, each side has an area of 16 square inches.

There are 4 sides and a bottom (no top) so the total area is 16 x 5 = 80 square inches.

If you were born before today's date, you would be 13; otherwise, you would still be 12.

The Pythagorean Theorem:  a² + b²  =  c²

So:                                       21² + b²  =  29²

Simplifying:                          441 + b²  =  841

Subtracting:                                   b²  =  400

Finding the square root:                  b  =  20                       (ignoring the negative answer)

1one bus has 28 kids thats the unit rate

-| 4x - 2 |  =  -10

Divide both sides by -1:     | 4x - 2 |  =  10

Because it's an absolute value problem:  Either  4x - 2  =  10  or  4x - 2  =  -10

If  4x - 2  =  10     --->     4x  =  12     --->     x  =  3

If  4x - 2  =  -10     --->     4x  =  -8     --->   x  =  -2

Okay, I thank you for drawing our attention to your question properly,  I have tried my best to answer.

The answers may not be correct. :)

Okay, you did not answer my question but I suppose it is same to assume you want to put the candy in the lunch boxes.

I have 3 pieces of candy to place in 4 lunch boxes.

In how many ways can I do this if:

(a) The candies are all different and the lunch boxes are all different?

Well you can put the first peice in any lunch box that is 4 choices, the same with the second 3rd so

it wll be   4*4*4 = 64 ways

(b) The candies are all the same and the lunch boxes are all the same?

1,1,1,0

2,1,0,0

3,0,0,0

That is 3 ways because the only thing that matters is how many lollies is in each box.

(c) The candies are all the same and the lunch boxes are all different?

3 from above * 4!  =    3*4*3*2*1    I think (not really sure )

(d) The candies are all different and the lunch boxes are all the same?

3*3!  maybe

(e) Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different?

3*4! * 3      I'm guessing.                      :))