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3 years ago

Please help me with this

Mathematics

1 answer:

STALIN [3.7K]3 years ago

4 0

Answer: none of these

Step-by-step explanation:

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Which table represents the graph of a logarithmic function in the form Y-log, when 5 > 1?

Viktor [21]

Answer:

Determine the domain and range of a logarithmic function.

Determine the x-intercept and vertical asymptote of a logarithmic function.

Identify whether a logarithmic function is increasing or decreasing and give the interval.

Identify the features of a logarithmic function that make it an inverse of an exponential function.

Graph horizontal and vertical shifts of logarithmic functions.

Graph stretches and compressions of logarithmic functions.

Graph reflections of logarithmic function

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

5. Find the surface area of the cone. Round to the nearest hundredth. 2 ft 6 ft

den301095 [7]

Answer:

$52.3m^2$

Step-by-step explanation:

Use these formulas

$A=\pi rl+\pi r^2\\l=\sqrt{r^2+h^2}$

Now solve

$A=\pi r(r+\sqrt{h^2+r^2} )\\=\pi *2*(2\sqrt{6^2+2^2} )\\=52.3m^2$

4 0

3 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago

A company is offering to pay a stadium $750,000 per year for naming rights. If the administrative costs for this sponsorship are

juin [17]

Answer: The costs are 8.4% of the total revenue.

The stadium is going to make $750,000 for the naming rights of the stadium. The cost to the stadium is on $63,000.

To find the percent, we have to divide the two values and multiply by 100.

63000 / 750000 x 100 = 8.4%

7 0

3 years ago

-6.3 as a mixed number

diamong [38]

Answer:

uit is the mixed number of decimal point

3 0

3 years ago