Hello :
f'(7) = the slope of the tangent line : let A(7,4) B(0,3)<span>
<span>the slope is : (YB - YA)/(XB -XA)= (3-4)/(0-7) = -1/-7 = 1/7=f'(7)
the equation of the tangent line is :
y-3 = 1/7 (x-0)
y = (1/7)x+3
the line tangent and the graph of : f </span></span><span>passes through the point A(7, 4)
x= 7 y=4 so : f(7) =4</span>
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:
hi there the answer is 72
Step-by-step explanation:
sorry but my work is on a piece of paper in front of me
Answer:
494235234934-1304039445
Step-by-step explanation:
4r567890e-r0987ytdghcn
b=5, commutative property
c=3, associative property
