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Yuliya22 [10]
3 years ago
15

629,999 rounded to the nearest ten thousand

Mathematics
1 answer:
Elis [28]3 years ago
8 0
It is 630,000
hope I helped:)
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Olegator [25]

Answer:

a) The Total fee, that Mike charges for 4 hours of instruction is $152

b) The total fee, that Tony charges for 4 hours of instruction is $148

c) The difference between total fee Mike and Tony charge for 4 hours of instructions is $4

Step-by-step explanation:

a) What is the total fee, that mike charges for 4 hours of instruction.

Mike charges $12 for travel expenses and $35 per hour for instruction.

So, the equation will be: y=35h+12 where h is number of hours and y is cost

Put h=4 and find cost

y=35h+12\\y=35(4)+12\\y=140+12\\y=152\\

So, the Total fee, that Mike charges for 4 hours of instruction is $152

b) What is the total fee, that Tony charges for 4 hours of instruction

Tony's fee can be modelled by C=32h+20 where h is number of hours and C is cost

Put h=4

C=32h+20\\C=32(4)+20\\C=128+20\\C=148

So, the total fee, that Tony charges for 4 hours of instruction is $148

c) What is the difference between total fee Mike and Tony charge for 4 hours of instructions.

Cost for Mike = $152

Cost for Tony = $148

Difference = 152 - 148 = 4

So, the difference between total fee Mike and Tony charge for 4 hours of instructions is $4

8 0
2 years ago
Let f(x) = -5e^(-x/3)<br><br> Find f^(6) (1)
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Let's take the first derivative:

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Notice that we can write this as:

f'(x) = -\dfrac{f(x)}{3}.

By taking the derivative of both sides n times, we get:

f^{(n+1)}(x) = -\dfrac{f^{(n)}}{3}.

This means that each time you take a derivative, a factor of -\dfrac{1}{3} will appear. So we conclude that:

f^{(n)}(x) = \left(-\dfrac{1}{3}\right)^nf(x).

Taking n=6 and x=1, we get:

f^{(6)}(1) = \left(-\dfrac{1}{3}\right)^6 f(1) = \dfrac{-5e^{-1/3}}{729} = -\dfrac{5}{729}e^{-1/3}.

So we finally get:

\boxed{-\dfrac{5}{729}e^{-1/3}}.

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