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4 years ago

Graph the linear equation. Find three

Mathematics

1 answer:

belka [17]4 years ago

7 0

Answer:

Part 1) The graph in the attached figure

Part 2) Three points that solve the equation could be (-18,0),(0,4.5) and (-2,4)

Step-by-step explanation:

we know that

To graph a line we need to plot two points

If a point solve the linear equation, then the point is a solution of the linear equation

step 1

Find the first point that solve the equation

Find the x-intercept

The x-intercept is the value of x when the value of y is equal to zero

For y=0

$x - 4(0) =-18$

$x=-18$

The x-intercept is the point (-18,0)

step 2

Find the second point that solve the equation

Find the y-intercept

The y-intercept is the value of y when the value of x is equal to zero

For x=0

$0 - 4y =-18$

$-4y=-18$

$y=4.5$

The y-intercept is the point (0,4.5)

step 3

With the intercepts graph the line

Plot the intercepts and join the points to graph the line

using a graphing tool

see the attached figure

step 4

Find the third point that solve the equation

Assume a value for x and then solve for y

I assume x=-2

substitute in the linear equation and solve for y

$-2 - 4y =-18$

$-4y=-18+2$

$-4y=-16$

$y=4$

The third point is the point (-2,4)

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3 years ago

The quotient of a number and - 2/3 is -9/10.

Ad libitum [116K]

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3 years ago

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For the following integral, find the approximate value of the integral with 4 subdivisions using midpoint, trapezoid, and Simpso

PIT_PIT [208]

Answer:

$\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}$

$\textsf{Trapezium rule}: \quad \pi$

$\textsf{Simpson's rule}: \quad \dfrac{4 \pi}{3}$

Step-by-step explanation:

Midpoint rule

$\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

Trapezium rule

$\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}$

Simpson's rule

$\displaystyle \int_{a}^{b} y \:\:\text{d}x \approx \dfrac{1}{3}h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{n-2}+4y_{n-1}+y_n\right)\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}$

Given definite integral:

$\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x$

Therefore:

a = 0
b = 2π

Calculate the subdivisions:

$\implies h=\dfrac{2 \pi - 0}{4}=\dfrac{1}{2}\pi$

Midpoint rule

Sub-intervals are:

$\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]$

The midpoints of these sub-intervals are:

$\dfrac{1}{4} \pi, \dfrac{3}{4} \pi, \dfrac{5}{4} \pi, \dfrac{7}{4} \pi$

Therefore:

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}$

Trapezium rule

$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} &&&&&\\ x & 0 & \dfrac{1}{2}\pi & \pi & \dfrac{3}{2} \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}$

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}$

Simpson's rule

$\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}$

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2 years ago

Use DeMoivre's Theorem to find (3cis*pi/6)^3

Sonbull [250]

The correct value of (3cis(pi/6))³ is 27i.

<h3>What is Complex Number?</h3>

Complex numbers are numbers that consist of two parts — a real number and an imaginary number. Complex numbers are the building blocks of more intricate math, such as algebra.

Given the complex number in polar coordinate expressed as

z = r(cos∅+isin∅)

zⁿ = {r(cos∅+isin∅)}ⁿ

According to DeMoivre’s Theorem;

zⁿ = rⁿ(cosn∅+isinn∅)

Given the complex number;

(3cis(pi/6))³

= {3(cosπ/6 + isinπ/6)}³

Using DeMoivre’s Theorem;

= 3³(cos3π/6 + isin3π/6)

= 3³(cosπ/2 + isinπ/2)

= 3³(0 + i(1))

= 27i

Thus, the correct value of (3cis(pi/6))³ is 27i.

Learn more about Complex number from:

brainly.com/question/10251853

#SPJ1

7 0

2 years ago