Step-by-step explanation:
This seems like you just want to figure out the circumference of the manhole cover. The formula for the circumference of a circle is pi (3.14) multiplied by the diameter (d) of the circle so, circumference=πd. (π is the symbol for pi and approx. equals 3.14)
Circumference = πd
= 3.14(d)
= 3.14(3)
= 9.42 ft.
The length of the brass grip-strip will be 9.42 ft.
If the problem was stated in terms of the radius of the manhole cover then the formula would be circumference = 2πr which is the radius multiplied by 2 then multiplied by pi.
The radius of a circle is the distance from the center to the edge and the diameter is the distance from one edge of the circle to the other passing through the center of the circle.
Well, if the grip strip were of no width and could be straightened out to a line (which a piece of rubber cut in a circle couldn't be), then the length of the grip would correspond to the circumference of the manhole cover.
Circumference = 2*PI*radius = PI*diameter so your answer is 3*PI feet long.
Answer:
first one with -1...21
Step-by-step explanation:
only one that follows quadratic formula format
Answer:
Step-by-step explanation:
Given:
elongation, x = 0.50 in
Force, f = 9000 lb
Young modulus, E = 10,000,000 psi
Maximum Stress, Sm = 30000 psi
Length, L = 16 ft
Converting ft to in,
12 in = 1 ft
=16 × 12 = 192 in
Young modulus, E = stress/strain
Stress = force/area, A
Strain = elongation, x/Length, L
E = f × L/A × E
1 × 10^7 = stress/(0.5/16)
= 26041.7 psi
Minimum stress = 26041.7 psi
Maximum stress = 30,000 psi
Stress = force/area
Area = 9000/26041.7
= 0.3456 in^2
Stress = force/area
Area = 9000/30000
= 0.3 in^2
Using minimum area of 0.3 in^2,
A = (pi/4)(d^2)
0.3 in^2 = (pi/4)(d^2)
d = 0.618 inches
diameter, d = 0.618 inches
Answer:
They lose about 2.79% in purchasing power.
Step-by-step explanation:
Whenever you're dealing with purchasing power and inflation, you need to carefully define what the reference is for any changes you might be talking about. Here, we take <em>purchasing power at the beginning of the year</em> as the reference. Since we don't know when the 6% year occurred relative to the year in which the saving balance was $200,000, we choose to deal primarily with percentages, rather than dollar amounts.
Each day, the account value is multiplied by (1 + 0.03/365), so at the end of the year the value is multiplied by about
... (1 +0.03/365)^365 ≈ 1.03045326
Something that had a cost of 1 at the beginning of the year will have a cost of 1.06 at the end of the year. A savings account value of 1 at the beginning of the year would purchase one whole item. At the end of the year, the value of the savings account will purchase ...
... 1.03045326 / 1.06 ≈ 0.9721 . . . items
That is, the loss of purchasing power is about ...
... 1 - 0.9721 = 2.79%
_____
If the account value is $200,000 at the beginning of the year in question, then the purchasing power <em>normalized to what it was at the beginning of the year</em> is now $194,425.14, about $5,574.85 less.
