Answer:
Okay . That looks interesting. Can't help.
Answer:
y = -5x + 20
Step-by-step explanation:
Lets start with the first given values for x and y
(2,10)
Lets try out the first equation
y = 5x + 4
y = 5(2) + 4
y = 14
This equation wouldn't work because the y value is not 10 when the x value is 2
y = 5x + 20
y = 5(2) + 20
y = 30
This equation doesn't work either
y = -5x + 20
y = -5(2) + 20
y = -10 + 20
y = 10
This would be the correct equation since your y value is 10 when the x value is 2
Answer:
Step-by-step explanation:
Given:
x = 2cost,
t = (1/2)arccosx
y = 2sint
dy/dx = dy/dt . dt/dx
dy/dt = 2cost
dt/dx = -1/√(1 - x²)
dy/dx = -2cost/√(1 - x²)
Differentiate again to obtain d²y/dx²
d²y/dx² = 2sint/√(1 - x²) - 2xcost/(1 - x²)^(-3/2)
At t = π/4, we have
(√2)/√(1 - x²) - (√2)x(1 - x²)^(3/2)
Answer:
if there can be no more than 5 students, then anything less than 5 students will suffice. However we don’t want any left over ones cause we don’t want that one kid sitting alone during lunch so we can simply put <u><em>4 into each group.</em></u>
Step-by-step explanation:
Hope this helped :DDD
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
