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dalvyx [7]
2 years ago
13

Find the equation of a line perpendicular to -x=4-y that passes through the point (-3,8)

Mathematics
1 answer:
Wewaii [24]2 years ago
8 0

9514 1404 393

Answer:

  y = -x +5

Step-by-step explanation:

Solving the given equation for y, we have ...

  y = x +4

The slope of this line is the coefficient of x: 1. The slope of the perpendicular line will be the opposite reciprocal of this: -1/1 = -1. The y-intercept of the perpendicular line can be found from ...

  b = y -mx = 8 -(-1)(-3) = 5

The perpendicular line has equation ...

  y = -x +5

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Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

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}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
What is the rule for finding the sum of two negative integers
ArbitrLikvidat [17]

adding two negative integers always yields a negative sum

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C. Both can be used to withdraw money from ATMs.

8 0
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How would I translate this algebraic expression?
FromTheMoon [43]

Answer: \frac{5}{7} (2+\frac{6}{r} )

Step-by-step explanation:

Since, the quotient of 6 and r = \frac{6}{r}

The sum of two and the quotient of 6 and r

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And,  Five-sevenths of the sum of two and the quotient of 6 and r

= \frac{5}{7} (2+\frac{6}{r} )

Thus, the required expression is, \frac{5}{7} (2+\frac{6}{r} )


3 0
3 years ago
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