Answer:
d=10u
Q(5/3,5/3,-19/3)
Step-by-step explanation:
The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane
, then r will have the next parametric equations:

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

Substitute the value of
in the parametric equations:

Those values are the coordinates of Q
Q(5/3,5/3,-19/3)
The distance from Po to the plane

125% my guys im sorry if its wrong :(
Answer:
15
Step-by-step explanation:
First we need to remove the parenthesis
4 + 37 + 8 - (34)
4 + 37 + 8 -34
Add 4 and 37
41 + 8 - 34
Add 41 and 8
49 - 34
Subtract
15
Well first do you parentheses and the multiply all of your exponets
Answer:
The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of
.
Step-by-step explanation:
The formula from the maximum distance of a projectile with initial height h=0, is:

Where
is the initial velocity.
In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is
. The critical points of the function are those who make
:


The critical value inside the interval is
.

The second step is to find the values of the function at the endpoints of the interval:

The biggest value of f is gived by
, therefore
is the absolute maximum.
In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of
.