Answer:
f(x) = 1500(0.97)ˣ
Step-by-step explanation:
Given that:
f(x) = abˣ
a is the initial bone density, that is the bone density at 0 years, x is the number of years and b is any real value. For an exponential growth, b > 1 while for an exponential decay, b < 1.
Since the bone has a current density of 1,500 kg/m³, hence a = 1500.
The density is lost at a rate of 3% annually, therefore b = 100% - 3% = 97% = 0.97.
Therefore substituting the values of a and b into the function gives:
f(x) = 1500(0.97)ˣ
The answer to 4067 / 56 is<span>72.625
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Answer:
6x+6y-5 distributive property
Step-by-step explanation:
6(x+y)-5
Distribute using the distributive property
6x+6y-5
Answer:
(- 1, 1 )
Step-by-step explanation:
x + 5y = 4 → (1)
4x + 12y = 8 → (2)
multiplying (1) by - 4 and adding to (2) will eliminate x
- 4x - 20y = - 16 → (3)
add (2) and (3) term by term to eliminate x
0 - 8y = - 8
- 8y = - 8 ( divide both sides by - 8 )
y = 1
substitute y = 1 into either of the 2 equations and solve for x
substituting into (1)
x + 5(1) = 4
x + 5 = 4 ( subtract 5 from both sides )
x = - 1
As a check
substitute these values into the left side of both equations and if equal to the right side then they are the solution.
- 1 + 5(1) = - 1 + 5 = 4 = right side
4(- 1) + 12(1) = - 4 + 12 = 8 = right side
(- 1, 1 ) is the solution to the system of equations
Answer:
Step-by-step explanation:
Let c represents child bikes and a represents adult bikes.
Given : Each child bike requires 4 hours to build and 4 hours to test. Each adult bike requires 6 hours to build and 4 hours to test.
With the number of workers, the company is able to have up to 120 hours of building time and 100 hours of testing time for a week.
Then, the required system of inequality :-
If company make 10 child bikes and 12 adult bikes in the week.
Then Put c=10 and a=12 bikes in (1) and (2).
⇒Bike order meets the restrictions
⇒Bike order meets the restrictions
Hence, the system of inequality best explains whether the company can build 10 child bikes and 12 adult bikes in the week.
