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
Answer:
one solution
Step-by-step explanation:
it has one solution because the slopes of the two equations(1x & 2x) are different, which means the lines of the two equations are not parallel, they have an intersection point(also means one solution).
Answer:
hope this helps and pls mark as brainliest
Answer:
m∠EGF = 65° and m∠CGF = 115°
Step-by-step explanation:
Given;
∠EFG = 50°
EF = FG
Solution,
In ΔEFG m∠EFG = 50° and EF = FG.
Since triangle is an isosceles triangle hence their base angles are always equal.
∴
Let the measure of ∠EGF be x.
∴ 
Now by angle Sum property which states "The sum of all the angles of a triangle is 180°."
m∠EFG + m∠FEG + ∠EGF = 180
Hence
m∠EGF = 65°
Also 'The sum of angles that are formed on a straight line is equal to 180°."
m∠EGF + m∠CGF = 180°
65° + m∠CGF = 180°
m∠CGF = 180° - 65° = 115°
Hence m∠EGF = 65° m∠CGF = 115°
