Answer:
Answer is "No real Solution"
Answer:
We multiply by 2 and add 4
Step-by-step explanation:
To find the pattern, we find the slope
m = ( y2-y1)/(x2-x1)
m = ( 10-6)/(3-1)
m = 4/2
m =2
We are multiplying by 2
y = mx+b where m is the slope and b is the y intercept
y = 2x+b
Using the point (1,6)
6 = 2(1)+b
6 = 2+b
The y intercept is 4
y = 2x+4
We multiply by 2 and add 4
U = ( -8 , -8)
v = (-1 , 2 )
<span>the magnitude of vector projection of u onto v =
</span><span>dot product of u and v over the magnitude of v = (u . v )/ ll v ll
</span>
<span>ll v ll = √(-1² + 2²) = √5
</span>
u . v = ( -8 , -8) . ( -1 , 2) = -8*-1+2*-8 = -8
∴ <span>(u . v )/ ll v ll = -8/√5</span>
∴ the vector projection of u onto v = [(u . v )/ ll v ll] * [<span>v/ ll v ll]
</span>
<span> = [-8/√5] * (-1,2)/√5 = ( 8/5 , -16/5 )
</span>
The other orthogonal component = u - ( 8/5 , -16/5 )
= (-8 , -8 ) - <span> ( 8/5 , -16/5 ) = (-48/5 , -24/5 )
</span>
So, u <span>as a sum of two orthogonal vectors will be
</span>
u = ( 8/5 , -16/5 ) + <span>(-48/5 , -24/5 )</span>
Polygon ABCD goes through a sequence of rigid transformations to form polygon A′B′C′D′. The sequence of transformations involved is a reflection across the _Y-axis _, followed by a reflection across the line _y=-x__.
There is a not so well-known theorem that solves this problem.
The theorem is stated as follows:
"Each angle bisector of a triangle divides the opposite side into segments proportional in length to the adjacent sides" (Coxeter & Greitzer)
This means that for a triangle ABC, where angle A has a bisector AD such that D is on the side BC, then
BD/DC=AB/AC
Here either
BD/DC=6/5=AB/AC, where AB=6.9,
then we solve for AC=AB*5/6=5.75,
or
BD/DC=6/5=AB/AC, where AC=6.9,
then we solve for AB=AC*6/5=8.28
Hence, the longest and shortest possible lengths of the third side are
8.28 and 5.75 units respectively.
