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3 years ago

How do you Graph y-3=1/3(x+1)

Mathematics

1 answer:

jekas [21]3 years ago

6 0

Answer:

You can find the point of intersection of the line with the y-axis and the the point of intersection of the line with the x-axis. See the graph attached.

Step-by-step explanation:

Find the intersection with the y-axis. Substitute $x=0$ into the equation and solve for y:

$y-3=\frac{1}{3}(x+1)\\\\y-3=\frac{1}{3}(0+1)\\\\y-3=\frac{1}{3}(1)\\y=\frac{1}{3}+3\\\\y=\frac{10}{3}$

$y$ ≈3.33

Find the intersection with the x-axis. Substitute $y=0$ into the equation and solve for x:

$y-3=\frac{1}{3}(x+1)$

$0-3=\frac{1}{3}(x+1)\\(-3)(3)=x+1\\-9-1=x\\x=-10$

Now, you know that the line passes through the point (0,3.33) and (-10,0). Now you can graph the function. (Observe the graph attached)

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Masteriza [31]

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2 years ago

What is the area of a garden that measures 4mx4mx6mx4mx10mx8m

topjm [15]

The area of the garden is 120 square metre

<h2>Explanation:</h2>

The picture of the garden is shown below. In this problem, we have:

One square
Two rectangles

Recall that the area of a square is given by:

$A_{s}=L^2 \\ \\ A_{s}:Area \ of \ a \ square \\ L:Side \ of \ the \ square$

On the other hand, the area of a rectangle is given by:

$A_{r}=L_{1}\times L_{2} \\ \\ \\ A_{r}:Area \ of \ a \ rectangle \\ \\ L_{1}:Side \ 1 \ of \ the \ rectangle \\ \\ L_{2}:Side \ 2 \ of \ the \ rectangle$

Therefore:

FOR THE SQUARE:

$L=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{s}=4^2 \\ \\ \boxed{A_{s}=16m^2}$

FOR THE RECTANGLE 1:

$L_{1}=6m \\ L_{2}=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{1}}=6(4) \\ \\ \boxed{A_{r_{1}}=24m^2}$

FOR THE RECTANGLE 2:

$L_{1}=10m \\ L_{2}=8m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{2}}=10(8) \\ \\ \boxed{A_{r_{2}}=80m^2}$

Finally, the area of a garden is the sum of the three areas:

$A_{total}=A_{s}+A_{r_{1}}+A_{r_{2}} \\ \\ A_{total}=16m^2+24m^2+80m^2 \\ \\ A_{total}=16+24+80 \\ \\ \boxed{A_{total}=120m^2}$

<h2>Learn more:</h2>

Area of a pizza: brainly.com/question/12878495

#LearnWithBrainly

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Alexxandr [17]

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