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Ratling [72]
4 years ago
10

Solve the system restrictions by graphing if there is no solution or an infinite number solutions so state you set notation to e

xpress solution sets.{ y=4x, y=-x+20}
what are the points on the graph

Mathematics
1 answer:
Bingel [31]4 years ago
7 0

Answer:

  (x, y) = (4, 16)

Step-by-step explanation:

The first equation is a proportion, so you know right away that (0, 0) is on the graph. Pick any value you like for x, and y will be 4 times that. So, for example, the point (5, 20) will also be on the graph of y=4x.

The second equation has a y-intercept of 20 and a slope of -1. So, however many unit to the right of the y-axis you choose to go, the y-value will be down that number of units from 20. In addition to (0, 20), the point (5, 15) will be on the graph of y = -x+20.

__

Personally, I like to use a graphing calculator to show the graphical solution.

Here, the solution is (x, y) = (4, 16).

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A rectangle, a triangle, and two congruent semicircles were used to form the figure shown. Rectangle 5cm,20cmcircle : radius 5cm
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Answer:

The area of a shape is the amount of space it occupies

Step 1:

We start by calculating the area of the rectangle using:

\begin{gathered} A_{rectangle}=l\times b \\  \end{gathered}

By substituting the values, we will have

\begin{gathered} A_{rectangle}=l\times b \\ A_{rectangle}=20cm\times5cm \\ A_{rectangle}=100cm^2 \\ A_1=100cm^2 \end{gathered}

Step 2:

we calculate the area of the triangle using:

A_{triangle}=\frac{1}{2}\times base\times height

By substituting the values, we will have

\begin{gathered} A_{tr\imaginaryI angle}=\frac{1}{2}\times base\times he\imaginaryI ght \\ A_{tr\mathrm{i}angle}=\frac{1}{2}\times10cm\times8cm \\ A_{tr\mathrm{i}angle}=\frac{80cm^2}{2} \\ A_{tr\mathrm{i}angle}=40cm^2 \\ A_2=40cm^2 \end{gathered}

Step 3:

Calculate the area of the two semicircles

A_{semicircle}=\frac{\pi r^2}{2}

By substituting the values, we will have

\begin{gathered} A_{sem\imaginaryI c\imaginaryI rcle}=\frac{\pi r^{2}}{2} \\ A_{sem\mathrm{i}c\mathrm{i}rcle}=3.14\times\frac{5^2}{2} \\ A_{sem\mathrm{i}c\mathrm{i}rcle}=39.25cm^2 \\ Area\text{ of two semicircle will be} \\ A_3=39.25cm^2\times2 \\ A_3=78.5cm^2 \end{gathered}

Step 4:

Calculate the area of the shape

We will calculate the area of the shape by adding all the individual areas together

A_{shape}=A_1+A_2+A_3

By substituting the values, we will have

\begin{gathered} A_{shape}=A_{1}+A_{2}+A_{3} \\ A_{shape}=100cm^2+40cm^2+78.5cm^2 \\ A_{shape}=218.5cm^3 \end{gathered}

Hence,

The area of the shape will be

\Rightarrow218.5cm^2

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