If 4x + y = 13 - y = 2 then y =
2 answers:
You might be interested in
Hi, We can to calculate the vectors.
And the determinant will be the plan Z
Let A = (0,03), B =(0,2,0) , C = (1,0,0) and D = (0,0,0)
Then,
AB = B - A
Replacing the points:
AB = (0,2,0) - (0,0,3)
AB = (0i, 2j , -3k)
----------------------------
Already the vector AC = C -A
That's is,
AC = (1,0,0) - (0,0,3)
AC = (1i, 0j, -3k)
Then,
The plan =![\left[\begin{array}{ccc}x&y&z\\0&2&-3\\1&0&-3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26y%26z%5C%5C0%262%26-3%5C%5C1%260%26-3%5Cend%7Barray%7D%5Cright%5D%20)
Solving it, we will have:
Plan: -6x -3y -2z + d = 0
Replacinng any point to find the value of d
Example the point A =(0,0,3)
-6(0) -3(0) -2(3) + d = 0
-6+d = 0
d = 6
Then, The us equation will stay of form following :
-6x -3y -2z +6 = 0
or
6x + 3y +2z -6 = 0
Isolating 2z:
2z = 6 -6x - 3y
Dividing both the sides od equation by 2
z = 3 - 3x - 3y/2
Then,
Now, Let's find the <span>domain in xy
</span>
|y
| (0,2)
|\
| \
| \
| \ (1,0)
------------------------- x
b = Cut in y
then b will be = 2
As y = ax + b
y = ax + 2
We have the point = (1,0)
Replace in the equation
0 = a(1) + 2
0 = a + 2
Isolate a
a = -2
Then us stay:
y = -2x + 2
-------------------------------------
With ,
----------------------------------------
Now putting 3x/2(-2x+2)² as commu factor
![\\ = \int\limits^1_0 {(\frac{3x(-2x+2)^2}{2} - \frac{3x^2(-2x+2)^2}{2} - \frac{3x(-2x+2)^3}{6} )} \, dx \\ \\ = \int\limits^1_0 { \frac{3x}{2}(-2x+2)^2[ 1- x- \frac{1}{3} (-2x+2)] } \, dx \\ \\ = \int\limits^1_0 { \frac{3x}{2}(-2x+2)^2[ 1- x+ \frac{2x}{3} - \frac{2}{3} ] } \, dx \\ \\ = \int\limits^1_0 { \frac{3x}{2}(-2x+2)^2[ \frac{1}{3} - \frac{x}{3}] } \, dx \\ \\ = \int\limits^1_0 { \frac{3x}{2}(-2x+2)^2( \frac{1-x}{3} ) } \, dx ](https://tex.z-dn.net/?f=%20%5C%5C%20%3D%20%20%5Cint%5Climits%5E1_0%20%7B%28%5Cfrac%7B3x%28-2x%2B2%29%5E2%7D%7B2%7D%20-%20%5Cfrac%7B3x%5E2%28-2x%2B2%29%5E2%7D%7B2%7D%20-%20%5Cfrac%7B3x%28-2x%2B2%29%5E3%7D%7B6%7D%20%29%7D%20%5C%2C%20dx%20%0A%20%5C%5C%20%0A%20%5C%5C%20%3D%20%20%5Cint%5Climits%5E1_0%20%7B%20%5Cfrac%7B3x%7D%7B2%7D%28-2x%2B2%29%5E2%5B%201-%20x-%20%5Cfrac%7B1%7D%7B3%7D%20%28-2x%2B2%29%5D%20%7D%20%5C%2C%20dx%20%0A%20%5C%5C%20%0A%20%5C%5C%20%3D%20%20%5Cint%5Climits%5E1_0%20%7B%20%5Cfrac%7B3x%7D%7B2%7D%28-2x%2B2%29%5E2%5B%201-%20x%2B%20%5Cfrac%7B2x%7D%7B3%7D%20-%20%5Cfrac%7B2%7D%7B3%7D%20%5D%20%7D%20%5C%2C%20dx%20%0A%20%5C%5C%20%0A%20%5C%5C%20%3D%20%20%5Cint%5Climits%5E1_0%20%7B%20%5Cfrac%7B3x%7D%7B2%7D%28-2x%2B2%29%5E2%5B%20%20%5Cfrac%7B1%7D%7B3%7D%20%20-%20%5Cfrac%7Bx%7D%7B3%7D%5D%20%7D%20%5C%2C%20dx%20%0A%20%5C%5C%20%0A%20%5C%5C%20%3D%20%20%5Cint%5Climits%5E1_0%20%7B%20%5Cfrac%7B3x%7D%7B2%7D%28-2x%2B2%29%5E2%28%20%5Cfrac%7B1-x%7D%7B3%7D%20%29%20%7D%20%5C%2C%20dx%20%0A%0A)
And the determinant will be the plan Z
Let A = (0,03), B =(0,2,0) , C = (1,0,0) and D = (0,0,0)
Then,
AB = B - A
Replacing the points:
AB = (0,2,0) - (0,0,3)
AB = (0i, 2j , -3k)
----------------------------
Already the vector AC = C -A
That's is,
AC = (1,0,0) - (0,0,3)
AC = (1i, 0j, -3k)
Then,
The plan =
Solving it, we will have:
Plan: -6x -3y -2z + d = 0
Replacinng any point to find the value of d
Example the point A =(0,0,3)
-6(0) -3(0) -2(3) + d = 0
-6+d = 0
d = 6
Then, The us equation will stay of form following :
-6x -3y -2z +6 = 0
or
6x + 3y +2z -6 = 0
Isolating 2z:
2z = 6 -6x - 3y
Dividing both the sides od equation by 2
z = 3 - 3x - 3y/2
Then,
Now, Let's find the <span>domain in xy
</span>
|y
| (0,2)
|\
| \
| \
| \ (1,0)
------------------------- x
b = Cut in y
then b will be = 2
As y = ax + b
y = ax + 2
We have the point = (1,0)
Replace in the equation
0 = a(1) + 2
0 = a + 2
Isolate a
a = -2
Then us stay:
y = -2x + 2
-------------------------------------
With ,
----------------------------------------
Now putting 3x/2(-2x+2)² as commu factor