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2 years ago

4x>=112 A.25 B.27 C.28 D.33 E.1 F.21

Mathematics

1 answer:

irina1246 [14]2 years ago

8 0

Answer:

Step-by-step explanation:

112 divided by 4 equals 28

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V144 = V121<br> What does this even mean

Over [174]

Answer:

That equation doesn't make sense, so this equation means "no solution"

Step-by-step explanation:

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3 years ago

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Which description best fits the graph

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Decreasing then increasing, y goes down then y goes back up

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3 years ago

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(100) points!!!! Help please

inysia [295]

Answer:

$\displaystyle a_8=142,11\ mg$

Step-by-step explanation:

Geometric sequence

Each term in a geometric sequence can be computed as the previous term by a constant number called the common ratio. The formula to get the term n is

$\displaystyle a_n=a_1r^{n-1}$

where $a_1$ is the first term of the sequence

The problem describes Georgie took 275 mg of the medicine for her cold in the first hour and that in each subsequent hour, the amount of medicine in her body is 91% (0.91) of the amount from the previous hour. It can be written as

amount in hour n = amount in hour n-1 * 0.91

This information provides the necessary data to write the general term as

$\displaystyle a_n=275\ .\ 0.91^{n-1}$

In the 8th hour (n=8), the remaining medicine present is Georgie's body is

$\displaystyle a_n=275\ .\ 0.91^{8-1}$

$\displaystyle a_n=275\ .\ 0.91^{7}$

$\boxed{\displaystyle a_8=142,11\ mg}$

6 0

3 years ago

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 $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

4 0

3 years ago

Which of these limits evaluate to 0?

Vikentia [17]

<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

$\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\$

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Part II

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\$

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Part III

$\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\$

7 0

2 years ago