Answer:
We have the system of equations:
y=1/3x+5
y=2/3x+5
To solve it graphically, we need to graph both lines and see in which point the lines intersect.
You can see the graph below, and you can see that the lines intersect in the point (0, 5)
Now, we can also solve this analytically.
We can use the fact that for the solution, we need y = y.
Then we can write:
(1/3)*x + 5 = (2/3)*x + 5
First, we can subtract 5 in both equations to get:
(1/3)*x = (2/3)*x
This only has a solution when x = 0.
Replacing x = 0 in one of the equations, we get:
y = (1/3)*0 + 5 = 5
Then the solution is x = 0, and y = 5, as we already could see in the graph.
Answer:
(3x+1)^2
Step-by-step explanation:
D) EFGH moved onto E'F'G'H after rotating 180 counterclockwise around the origin and the reflecting across the y-axis.
<h3>How to carry out transformations?</h3>
From online resources gotten about this question, for quadrilateral EFGH and quadrilateral E'F'G'H to be congruent, what we must do first is to rotate 180° counterclockwise around the origin and then move EFGH onto E'F'G'H'.
The last step to get this proof of congruency is to reflect across the y-axis.
Read more about transformations at; brainly.com/question/4289712
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Answer:
10m x 15m
Step-by-step explanation:
You are given some information.
1. The area of the garden: A₁ = 150m²
2. The area of the path: A₂ = 186m²
3. The width of the path: 3m
If the garden has width w and length l, the area of the garden is:
(1) A₁ = l * w
The area of the path is given by:
(2) A₂ = 3l + 3l + 3w + 3w + 4*3*3 = 6l + 6w + 36
Multiplying (2) with l gives:
(3) A₂l = 6l² + 6lw + 36l
Replacing l*w in (3) with A₁ from (1):
(4) A₂l = 6l² + 6A₁ + 36l
Combining:
(5) 6l² + (36 - A₂)l +6A₁ = 0
Simplifying:
(6) l² - 25l + 150 = 0
This equation can be factored:
(7) (l - 10)*(l - 15) = 0
Solving for l we get 2 solutions:
l₁ = 10, l₂ = 15
Using (1) to find w:
w₁ = 15, w₂ = 10
The two solutions are equivalent. The garden has dimensions 10m and 15m.
