The x-intercept is present where y = 0
2x + 5y - 10 = 0
2x - 10 = -5y
2x - 10 = -5(0)
2x = 10
x = 5
The y-intercept is present where x = 0
2x + 5y = 10
2x + 5y - 10 = 0
5y - 10 = -2(0)
5y = 10
y = 2
57.63 or 57.6
This works because you get 55 and then you’d get 84 so you’d multiply by 7. This would be 7 and you would put a decimal and carry a 0 so it would 70. So you’d end up with 57.6
Solving the given inequality for d, we get...
6d + 15 < 50
6d + 15-15 < 50-15 <<-- subtract 15 from both sides
6d < 35
6d/6 < 35/6 <<--- divide both sides by 6
d < 5.83
Which means that d can be any of the values in this set: {0, 1, 2, 3, 4, 5}
The smallest d can be is 0. In this scenario, Jeremy pays the $15 registration but doesn't rent the camera at all
The largest d can be is 5. In this scenario, Jeremy rents the camera for 5 days
Any larger value of d is not allowed as it would make the total cost go over $50
Notice how I'm rounding down regardless how close 5.83 is to 6
Answer:
(a) A = (20mg)/(2^(t/30))
(b) 12.6mg
(c) 129.6years
Step-by-step explanation:
To calculate the amount remaining after a number of half-lives, n, we can make use of:
Where A = amount remaining
B = initial amount
(a) A = (20mg)/(2^(t/30))
(b) Mass after 20years
A = (20mg)/(2^(20/30)) ≈ 12.6mg
(c) After how long will only 1mg remain:
1mg = (20mg)/(2^(t/30))
Taking log of both sides we have:
Log(20) = (t/30)log(2)
t/30 = (log(20))/(log(2)) ≈ 4.3
t/30 = 4.3
t = 30 x 4.3 ≈ 129.6years.
Answer:
Its out of A and C srry couldn't help much
