Find the linearization Upper L (x comma y )of the function f (x comma y )at Upper P 0. Then find an upper bound for the magnitud
e StartAbsoluteValue Upper E EndAbsoluteValue of the error in the approximation f (x comma y )almost equalsUpper L (x comma y )over the rectangle R. f (x comma y )equals3 ln x plus 5 ln y at Upper P 0 (1 comma 1 )R: StartAbsoluteValue x minus 1 EndAbsoluteValue less than or equals 0.4, StartAbsoluteValue y minus 1 EndAbsoluteValue less than or equals 0.3
1 answer:
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Answer: 1, 2, 5, and 10
Step-by-step explanation: To find the factors of 10, begin by dividing 10 by 1 which gives us 10. This tells us that 1 and 10 are factors of 10. Next, divide 10 by 2 which gives us 5. This tells us that 2 and 5 are also factors of 10. Next, divide 10 by 3. Notice that 10 doesn't divide evenly by 3 so 3 is not a factor of 10. Next, divide 10 by 4. Since 10 doesn't divide evenly by 4, then 4 is not a factor of 10. Next, divide 10 by 5 which gives us 2. This tells us that 5 and 2 are factors of 10 but notice that we already know that 5 and 2 are factors of 10. This means that 5 and 2 are repeat factors.
There is no need to repeat factors that we already know. In fact, as we are writing the factors of a number, it's important to understand that once we hit repeat factors, all other factors that we try will also repeat.
So here, dividing by anything greater than 5 will give us factors that we already know so we can stop here.
This means that the factors of 10 are 1, 2, 5, and 10.
Answer:
Approx 2.61 years.
Step-by-step explanation:
Given that you decide to put $5,000 in a savings account to save for a $6,000 down payment on a new car.
That means A=6000 and P=5000
If the account has an interest rate of 7% per year and is compounded monthly,
Then r=7%= 0.07 and n=12 (monthly)
Now we need to find about how long does it take until you have $6,000 without depositing any additional funds.
So let's plug these values into compound interest formula and fine time t
\left(1.00583333333333\right)[/tex]
Hence answer is approx 2.61 years.