To convert 8 kiloliters to liters, the first ratio is 1 kL/1,000 L. To set up the proportion, the second ratio must be _____.

Mathematics

2 answers:

dlinn [17]3 years ago

8 0

Answer:

B. $\frac{8kL}{x}$

Step-by-step explanation:

For 2 ratios to be the same, they have to be equal to one another, in this case, meaning if a part of one of the ratios is unknown, you must be able to solve it by setting the two ratios equal to one another:

$\frac{1kL}{1000L} =\frac{8kL}{x}\\x=\frac{8kL}{\frac{1kL}{1000L}} \\x=8kL*\frac{1000L}{1kL} \\x=8000L$

My name is Ann [436]3 years ago

6 0

It would be b. the first ratio is
1kL
----
1000

so you would match up the kL and not know how many regular liters there are.

8kL
-----
x

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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$