4(2 - x) > -2x - 3(4x + 1)
8 - 4x > -2x - 12x - 3
-4x + 2x + 12x > -3 - 8
10x > -11
x > -11/10
x > -1.1
Therefore, x = 0 and x = 10 zre solutions to the inequality.
Answer:
t(d) = 0.01cos(5π(d-0.3)/3)
Step-by-step explanation:
Since we are given the location of a maximum, it is convenient to use a cosine function to model the torque. The horizontal offset of the function will be 0.3 m, and the horizontal scaling will be such that one period is 1.2 m. The amplitude is given as 0.01 Nm.
The general form is ...
torque = amplitude × cos(2π(d -horizontal offset)/(horizontal scale factor))
We note that 2π/1.2 = 5π/3. Filling in the given values, we have ...
t(d) = 0.01·cos(5(d -0.3)/3)
.04b is 2% of 2b because .04 multiplied by 100 is 4 then you divide by 2 which gets you 2.
.04/2=x/100
cross multiply and divide
it is 98% smaller than 2b.
.2b is 10% of 2b because .2 multiplied by 100 is 20 then you divide by 2 and get 10.
.2/2=x/100
it is 90% smaller than 2b.
.56b is 28% of 2b because .56 multiplied by 100 is 56 then divide by 2 and you get 28.
.56/2=x/100
it is 72% smaller than 2b.
1.8b is 90% of 2b because 1.8 multiplied by 100 is 180 then divide it by 2. you get 90.
1.8/2=x/100
it is 10% smaller than 2b.
2.5b is 125% of 2b because 2.5 multiplied by 100 Is 250 and 250 divided by 2 is 125.
2.5/2=x/100
it is 25% larger than 2b
3b is 150% of 2b because 3 multiplied by 100 is 300 and 300 divided by 2 is 150.
3/2=x/100
it is 50% larger than 2b
For this case we have the following equation:

Where,
w: The weight of a spring in pounds
E: the energy stored by the spring in joules.
Substituting values we have:

Making the corresponding calculation:
Answer:
the approximate weight of the spring in pounds is:
Answer:
8 helicopters at most
Step-by-step explanation:
I worked on this problem and that was the answer not 99 like the other.