A machining company manufactures washers. The dimensions of the washer are 2cm for the inner circle and 4cm for the outer circle. And the inner and outer radii is positive and negative 2. The range of acceptable values is called the tolerance.
A. Write compound inequalities for the acceptance dimensions for the inner radius r and outer radius R.
B. Write an equation for the area of the face of the washer in terms of r,R, and pie.
C. Write a compound inequality for the area, to the nearest tenth of a square centimeter, or the face of the washer. HELP!!!!!!!
A machining company manufactures washers. The dimensions of the washer are shown in the figure, and the inner and outer radii can be +/-2% of the indicated values. This range of acceptable values is called the tolerance.
A) Write compound inequalities for the acceptable dimensions for the inner radius r and outer radius R.
B) Write an eqaution for the area of the face of the washer in terms of r, R, and pi.
C) Write a compound inequality for the area, to the nearest tenth of a square centimeter, of the face of the washer. (hint:Find the smallest possible area and the largest are and the largest possible area.)
PS: I really only need help on b and c. Please help
A. To write compound inequalities for the acceptance dimensions of the inner radius r and outer radius R, we need to consider the given dimensions and tolerances.
Since the inner radius of the washer is 2 cm with a positive and negative tolerance of 2 cm, we can write the compound inequality as:
-4 ≤ r ≤ 8
Similarly, since the outer radius of the washer is 4 cm with a positive and negative tolerance of 2 cm, we can write the compound inequality as:
2 ≤ R ≤ 6
B. To write an equation for the area of the face of the washer in terms of r, R, and π (pi), we can use the formula for the area of a washer, which is the difference between the areas of the outer and inner circles:
Area = π(R^2 - r^2)
C. To write a compound inequality for the area of the face of the washer, we need to consider the acceptable range of values for the area. Since the tolerance is not mentioned for the area, we assume that the acceptable range for the area is within the range of acceptable values for the inner and outer radii.
Using the compound inequalities we wrote in part A, we can substitute the values in the equation for the area:
π(2^2 - (-4)^2) ≤ Area ≤ π(6^2 - 8^2)
Simplifying this further:
π(4 - 16) ≤ Area ≤ π(36 - 64)
π(-12) ≤ Area ≤ π(-28)
Since the area cannot be negative, the lower bound of the compound inequality for the area becomes 0:
0 ≤ Area ≤ π(-28)
Therefore, the compound inequality for the area of the face of the washer, to the nearest tenth of a square centimeter, is:
0 ≤ Area ≤ -87.9646 (rounded to the nearest tenth)
I assume you meant that the tolerance is ±2mm or ±0.2cm
So, that means that
2.0-0.2 <= r <= 2.0+0.2
or
1.8 <= r <= 2.2
If I got that wrong, you can adjust the values.
Now you can probably handle the rest ...
and that's pi, NOT pie!
pie is what you eat.