Answer:
I'm not an expert here, this is a best guess!
But I would say if there is no chance that of him incurring excess costs of less than $500, then he knows without insurance he'll end up paying at least $500, possibly more out of pocket, without the insurance.
so I would say He ends up spending the least amount out if pocket by going with option A. for $75. that's $75 out of pocket with no deductible and it covers his $500+ in excess costs....B and C would also cover the excess, but would each cost $140 or $275 out of pocket at the end of the day....
with that being said, I'd say it's worth it to buy the insurance....even if he doesn't have any excess costs, he's spent $75 dollars for the peace of mind to know he's covered either way, and if he does incur the excess costs he's spent $75 rather that $500+....Even if the excess charges are only $100, which it says there is no chance of happening, but still, then he's still saved $25 altogether. Unless I'm reading it wrong, Option A saves him the most money either way, and is worth it to buy the insurance!
Answer:
2
Step-by-step explanation:
So we have the equation:
This is in the format point-slope form, where:
Here, m is the slope.
In our original equation, 2 replaces m.
Therefore, our slope is 2.
Answer:
5
Step-by-step explanation:
( divide numerator and denominator by 5 ) , then
= 
( with missing number ? = 5 )
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
Step-by-step explanation:
F(n)=|sin(n)|+|sin(n+1)|
then
F(n+π)=|sin(n+π)|+|sin(n+π+1)|=|sin(n)|+|sin(n+1)|=F(n)
and
F(π−n)=|sin(π−n)|+|sin(π−n+1)|=|sinn|+|sin(n−1)|≠F(n)
so we must prove when n∈(0,π), have
F(n)>2sin12
when n∈(0,π−1),then
F(n)=sinn+sin(n+1)=sinn(1+cos1)+sin1cosn
and n∈(π−1,π),then
F(n)=sinn−sin(n+1)
How prove it this two case have F(n)>2sin12? Thank you
and I know this well know inequality
|sinx|+|sin(x+1)|+|sin(x−1)|≥2sin1,x∈R
