Let y = √4+7t
then u= 4+7t
y=√u = u^½
du/dt= 7
dy/du = ½U^-½
dy/dt = du/dt • dy/du
= 7×½U^-½
= 7/2√U
= 7 / (2{√4+7t})
Almost got it!
x + 3 = 3(y + 2)/2 [Multiplied both sides by 3]
so x + 3 = (3y + 6)/2
2(x + 3) = 3y + 6 [Multiplied both sides by 2]
2x + 6 = 3y + 6
2x = 3y [Subtracted 6 from both sides]
x = 3y/2 [Divided both sides by 2]
x/3 = y/2 [Divided both sides by 3]
So you wrote y/3 instead of y/2
Hope this helped!
With a given parallel line and a given point on the line
we can use the point-line method: y-y0=m(x-x0)
where
y=mx+k is the given line, and
(x0,y0) is the given point.
Here
m=-10, k=-5, (x0,y0)=(-3,5)
=> the required line L is given by:
L: y-5=-10(x-(-3))
on simplification
L: y=-10x-30+5
L: y=-10x-25
It's C x5 the correct answer I'm not sure
Answer: u= 3.2
Step-by-step explanation:
divide both by 4
