A machinist has to manufacture a circular metal disk with area 1060π sq. cms. How close to the exact radius must the machinist control the radius if he is allowed an error tolerance of ±20π sq. cms. in the area of disk?

The radius of the disk must be close to 33cm

Step-by-step explanation:

Formula for calculating the area of a circular metal disk is Πr² since area of a circle is Πr². If the area of the disk is 1060π sq. cms, to get the radius of the disc, we will use the formula;

A = Πr² where;

A is the area of the disc

r is the radius of the disc

Given A = 1060π sq. cms

1060Π = Πr²

1060 = r²

r = √1060

r = 32.56cm approximately

If he is allowed a tolerance of ±20Πsq. cms in area, this means that we can use 1060±20Πsq. cms as the area which is equivalent to either 1080sq. cms or 1040sq. cms

If A = 1080sq. cms

r = √1080 = 32.86cm

If A = 1040sq. cms

r = √1040 = 32.25cm

If the values of the radius are compared, the radius of the metal disk must be close to 33cm approximately which are within the tolerance limit of the area.

c > -10.55

step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: