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3 years ago

Complete the table Due in 12 minutes

Mathematics

2 answers:

ikadub [295]3 years ago

8 0

Answer:

Step-by-step explanation:

the answer is the third one

because x is multiplying by 4

please mark me as brainliest

Mekhanik [1.2K]3 years ago

5 0

Answer:

4th one

when x goes up y=+4

so if x is 5 y is 20

but if y is 28 x is 7

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Draw a line representing the "run" and a line representing the "rise" of the line. State the slope of the line in simplest form.

Shtirlitz [24]

Answer:

$\displaystyle 1\frac{1}{3}$

Step-by-step explanation:

[3, 1] and [0, −3]

$\displaystyle \frac{-y_1 + y_2}{-x_1 + x_2} = m \\ \\ \frac{3 + 1}{0 + 3} = \frac{4}{3} = 1\frac{1}{3}$

I am joyous to assist you anytime.

4 0

3 years ago

Read 2 more answers

Write the equation of line whose b= -1 and m=5

vovangra [49]

Answer:

Step-by-step explanation:

y=mx+b

y=5x-1

4 0

3 years ago

Read 2 more answers

1/x-2 graphed<br><br><img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx-2%7D" id="TexFormula1" title="\frac{1}{x-2}" alt="\fra

bearhunter [10]

Answer:

Step-by-step explanation:

The given equation is expressed as

y = 1/(x - 2)

We will assume values for x and determine the corresponding values for y. Therefore,

When x = 0, y = 1/(0 - 2) = - 1/2

When x = 1, y = 1/(1 - 2) = - 1

When x = 2, y = 1/(2 - 2) = 0

When x = 3, y = 1/(3 - 2) = 1

When x = 4, y = 1/(4 - 2) = 1/2

To plot the graph, we would choose a suitable scale for the x and y axis of the graph.

Let 2 cm represent 1 unit on the vertical axis.

Let 2 cm represent 0.5 unit on the x axis.

The graph is shown in the attached photo

4 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

Julian sees a common denominator for 3 x 4 + 2

Rasek [7]

the denominator would be 12

7 0

3 years ago