Answer:
See proof below
Step-by-step explanation:
Assume that V is a vector space over the field F (take F=R,C if you prefer).
Let 
. Then, we can write x as a linear combination of elements of s1, that is, there exist 
and 
such that 
. Now, 
then for all 
we have that 
. In particular, taking 
with 
we have that 
. Then, x is a linear combination of vectors in S2, therefore 
. We conclude that 
.
If, additionally 
then reversing the roles of S1 and S2 in the previous proof, 
. Then 
, therefore 
.
It depends on which variable is eliminated.
If you multiply the second equation by -1 and eliminate the y:
3x - y = 7
-6x + y = -10
-3x = -3 or x = 1
If you multiply the first equation by -2 and eliminate the x:
-6x + 2y = -14
6x - y = 10
y = -4
![\bf \begin{cases} x=3\implies &x-3=0\\ x=1+3i\implies &x-1-3i=0\\ x=1-3i\implies &x-1+3i=0 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (x-3)(x-1-3i)(x-1+3i)=0 \\\\\\ (x-3)\underset{\textit{difference of squares}}{([x-1]-3i)([x-1]+3i)}=0\implies (x-3)([x-1]^2-[3i]^2)=0 \\\\\\ (x-3)([x^2-2x+1]-[3^2i^2])=0\implies (x-3)([x^2-2x+1]-[9(-1)])=0](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%3D3%5Cimplies%20%26x-3%3D0%5C%5C%20x%3D1%2B3i%5Cimplies%20%26x-1-3i%3D0%5C%5C%20x%3D1-3i%5Cimplies%20%26x-1%2B3i%3D0%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28x-3%29%28x-1-3i%29%28x-1%2B3i%29%3D0%20%5C%5C%5C%5C%5C%5C%20%28x-3%29%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%28%5Bx-1%5D-3i%29%28%5Bx-1%5D%2B3i%29%7D%3D0%5Cimplies%20%28x-3%29%28%5Bx-1%5D%5E2-%5B3i%5D%5E2%29%3D0%20%5C%5C%5C%5C%5C%5C%20%28x-3%29%28%5Bx%5E2-2x%2B1%5D-%5B3%5E2i%5E2%5D%29%3D0%5Cimplies%20%28x-3%29%28%5Bx%5E2-2x%2B1%5D-%5B9%28-1%29%5D%29%3D0)
[ correction added, Thanks to @stef68 ]
![\bf (x-3)([x^2-2x+1]+9)=0\implies (x-3)(x^2-2x+10)=0 \\\\\\ x^3-2x^2+10x-3x^2+6x-30=0\implies x^3-5x^2+16x-30=f(x) \\\\\\ \stackrel{\textit{applying a translation with a -2f(x)}}{-2(x^3-5x^2+16x-30)=f(x)}\implies -2x^3+10x^2-32x+60=f(x)](https://tex.z-dn.net/?f=%5Cbf%20%28x-3%29%28%5Bx%5E2-2x%2B1%5D%2B9%29%3D0%5Cimplies%20%28x-3%29%28x%5E2-2x%2B10%29%3D0%20%5C%5C%5C%5C%5C%5C%20x%5E3-2x%5E2%2B10x-3x%5E2%2B6x-30%3D0%5Cimplies%20x%5E3-5x%5E2%2B16x-30%3Df%28x%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bapplying%20a%20translation%20with%20a%20-2f%28x%29%7D%7D%7B-2%28x%5E3-5x%5E2%2B16x-30%29%3Df%28x%29%7D%5Cimplies%20-2x%5E3%2B10x%5E2-32x%2B60%3Df%28x%29)
Answer:
x = 18
18√2
Step-by-step explanation:
This is a special right triangle with angle measures 45° 45° 90°
in this special triangle the side lengths are expressed as a, a, and a√2
So x = a and x = 18
r = a√2 and r = 18√2
Answer:
2g² + 10g
Step-by-step explanation:
Let the,
Length of the rectangle be " l ".
Base of the rectangle be " b ".
l = 2g
b = g + 5
Formula : -
Area of the rectangle = lb
Area of the rectangle
= 2g ( g + 5 )
= 2g ( g ) + 2g ( 5 )
= 2g² + 10g
Therefore,
the area of a rectangle whose sides measure 2g and ( g + 5 ) is 2g² + 10g.
