One way in which to do this problem would involve subtracting 5 from 7 (result: 2) and then subtracting 3/5 from 8/9.
To subtract 3/5 from 8/9, you'd need to find the lowest common denominator (LCD) of 3/5 and 8/9, convert both fractions to have this LCD, and then subtract.
The LCD is (5)(9)=45. Then 8/9 and 3/5 become 40/45 and 27/45.
Subtracting 27/45 from 40/45 results in the fraction 13/45.
Then the full solution is 2 13/45.
You could also do this problem by converting 7 8/9 and 5 3/5 into improper fractions:
71/9 - 28/5. Again, the LCD is 45. Can you rewrite both fractions with 45 as the common denominator and then perform the subtraction?
The lengths of a right triangle's three sides can be expressed as 24, 32, and 40.
Let's work our way through the solution. According to the right triangle formula, a triangle's hypotenuse square equals the sum of its base square and its altitude square.
How to determine a right triangle's sides?
If leg an is absent, change the equation to its form when leg an is present on one side and compute the square root: a = (c2 - b2).
Leg b must be unknown otherwise. b = √(c² - a²)
The equation for hypotenuse c is: c = (a2 + b2)
to learn more about right triangle's sides refer to:
brainly.com/question/3223211
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Answer:A
Step-by-step explanation:
She is taxed at a rate of 2.9%, so each year she is taxed 213000 * 2.9% . calculating this, we find that every year, she must pay 6177 in taxes. However, we need the amount in a month, so we divide by 12, to get 6177/12= 514.75, or A
Answer:
We want to find:
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D)
Here we can use Stirling's approximation, which says that for large values of n, we get:
Because here we are taking the limit when n tends to infinity, we can use this approximation.
Then we get.
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7B%5Csqrt%7B2%2A%5Cpi%2An%7D%20%2A%28%5Cfrac%7Bn%7D%7Be%7D%20%29%5En%7D%20%7D%7Bn%7D%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7Bn%7D%7Be%2An%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D)
Now we can just simplify this, so we get:
![\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7Be%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D%20%5C%5C)
And we can rewrite it as:
The important part here is the exponent, as n tends to infinite, the exponent tends to zero.
Thus:
