Answer:
The shape is not there so i can not solve it
Step-by-step explanation:
3.) 5x-6=0, Add 6 to either side -> 5x=6, divide each side by 5 -> x=6/5 ---- 4.) x+y/3=5, multiple each side by 3 -> x+y=15, move y to the other side -> x=-y+15
Let no red ribbons=x
no. Of blue ribbons=x+3
No. Of yellow ribb9ns=x+7
Totsl ribbons=28
28=x+x+3+x+7
28=3x+10
3x=28-10
3x=18
X=18/3
X=6
No .of red ribbons=6
No. Of blue ribbons=x+3=9
No. Of yellow ribbons=x+7=13
How do you want this solved?
In the
plane, we have
everywhere. So in the equation of the sphere, we have

which is a circle centered at (2, -10, 0) of radius 4.
In the
plane, we have
, which gives

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.
In the
plane,
, so

which is a circle centered at (0, -10, 3) of radius
.