Answer:
0.000000001
Hope This Helps! Have A Nice Day!!
Answer:
C
Step-by-step explanation:
It usually works best to use the polynomial with fewer terms as the multiplier. A row of partial products is written for each term of the multiplier, so the fewer terms will result in fewer rows of partial products.
In order to keep like terms together, it is preferable to allocate a separate column of the multiplication tableau to each power of the operands or product. This means we want to make note of the fact that the cubic multiplicand has a coefficient of 0 for its x^2 term.
The best setup is the one shown in the attachment.
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>
Answer:
3 blocks east
5 blocks south
Step-by-step explanation:
Hope it helps
Subtract the x on the left from the 2x on the right side
