The line passes through (9,8) and is perpendicular to x-2y=-16

Mathematics

1 answer:

pickupchik [31]3 years ago

4 0

The equation of line perpendicular to x-2y=-16 passing through (9,8) is: y=-2x+26

Step-by-step explanation:

Given

$x-2y=-16\\Adding\ 2y\ on\ both\ sides\\x=2y-16\\Adding\ 16\ on\ both\ sides\\x+16=2y-16+16\\x+16=2y\\2y=x+16\\Dividing\ both\ sides\ by\ 2\\\frac{2y}{2}=\frac{x+16}{2}\\y=\frac{x}{2}+\frac{16}{2}\\y=\frac{1}{2}x+8\\$

The equation is in slope-intercept form, the coefficient of x will be the slope of given line. The slope is: 1/2

As the product of slopes of two perpendicular lines is -1.

$\frac{1}{2}*m=-1\\m=-1*\frac{2}{1}\\m=-2$

Slope intercept form is:

$y=mx+b$

Putting the value of slope

y=-2x+b

To find the value of b, putting (9,8) in the equation

$8=-2(9)+b\\8=-18+b\\b=8+18\\b=26$

Putting the values of b and m

$y=-2x+26$

Hence,

The equation of line perpendicular to x-2y=-16 passing through (9,8) is: y=-2x+26

Keywords: Equation of line, Slope-intercept form

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