Steps:
6 1/4 - 4 3/15 can also be written as 25/4 - 23/5
The LCD of 4 and 5 is 20.
That gives us 125/20 - 92/20.
Now, just subtract the numerators (leave the denominators the same).
That brings us to 33/20, which can be simplified to 1 13/20
Final Answer:
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<u><em>*Please mark brainliest if that helped!*</em></u>
Answer:
no
Step-by-step explanation:
5x would be multiplying 5 and x. So, if x=4, then 5x would be 20, and 20+10=30. But if you did 10x+5, it would be 40+5 which is 45. Therefore, the answers would not be the same.
The answer is true.
For example lets say the numbers given are 1, 5, 9, ...
To find the common difference, you subtract the 1st term from the 2nd term. In this case, we would do 5 - 1, which equals 4.
We must add the common difference to the last term in order to find the next term. Since the common difference is 4, we add 4 to 9, which makes 13. This applies only in the arithmetic sequence.
21 is the first one, 2.27 is the second answer. Hope it helps!
Answer:
4.45 ccs will remain after 5 hours.
Step-by-step explanation:
The anti-biotic in his body decreases at a rate proportional to the amount, Q(t), present at time t.
(dQ/dt) = - kQ ((Minus sign because it's a rate of reduction)
(dQ/dt) = -kQ
(dQ/Q) = -kdt
∫ (dQ/Q) = -k ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from Q₀ to Q and the Right hand side from 0 to t.
We obtain
In (Q/Q₀) = -kt
(Q/Q₀) = e⁻ᵏᵗ
Q(t) = Q₀ e⁻ᵏᵗ
the initial injection was 8 ccs and 5 ccs remain after 4 hours
Q₀ = 8 ccs,
At t = 4 hours, Q = 5 ccs
5 = 8 e⁻ᵏᵗ
e⁻ᵏᵗ = 0.625
-kt = In (0.625) = -0.47
-4k = 0.47
k = 0.1175 /hour
Q(t) = Q₀ e⁻⁰•¹¹⁷⁵ᵗ
At t = 5 hours, Q = ?
Q = 8 e⁻⁰•¹¹⁷⁵ᵗ
0.1175 × 5 = 0.5875
Q = 8 e(^-0.5875)
Q = 4.45 ccs
Hope this Helps!!!
