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3 years ago

Rounding 8,469 to one decimal place

Mathematics

1 answer:

wlad13 [49]3 years ago

3 0

Answer:

do you mean "8.469"? of so 8.5

Step-by-step explanation:

assuming the comma should be a decimal

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What is the value of AAA when we rewrite \left(\dfrac {6}{17}\right)^{9x}( 17 6 ) 9x (, start fraction, 6, divided by, 17, end

sineoko [7]

We have been given an expression $\left(\dfrac {6}{17}\right)^{9x}$ . We are asked to find the value of A when rewrite our given expression as $A^{x}$ .

To solve our given problem, we will use exponent properties.

Using exponent property $a^{mn}=(a^m)^n$ , we can rewrite our given expression as:

$\left(\dfrac {6}{17}\right)^{9x}=\left(\left(\dfrac {6}{17}\right)^9\right)^{x}$

Now, we will compare our expression with $A^{x}$ .

Upon comparing $\left(\left(\dfrac {6}{17}\right)^9\right)^{x}$ with $A^{x}$ , we can see that $A=\left(\dfrac {6}{17}\right)^9$ .

Therefore, the value of A is $\left(\dfrac {6}{17}\right)^9$ .

We can further simplify $\left(\dfrac {6}{17}\right)^9$ as:

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3 years ago

(a) Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 Correct

Luba_88 [7]

Splitting up the interval of integration into $n$ subintervals gives the partition

$\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]$

Each subinterval has length $\dfrac{1-0}n=\dfrac1n$ . The right endpoints of each subinterval follow the sequence

$r_i=\dfrac in$

with $i=1,2,3,\ldots,n$ . Then the left-endpoint Riemann sum that approximates the definite integral is

$\displaystyle\sum_{i=1}^n\frac{{r_i}^3}n$

and taking the limit as $n\to\infty$ gives the area exactly. We have

$\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac in\right)^3=\lim_{n\to\infty}\frac{n^2(n+1)^2}{4n^3}=\boxed{\frac14}$

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3 years ago