Answer:
Answer is 60%
Step-by-step explanation:
I hope it's helpful!
To find the point of intersection, we want to set the two equations equal to each other to find where they meet. The problem is, we have two variables, which means we can't just set them equal to each other as is. We need to manipulate the equations so that we can remove one of the variables at a time to solve for the other one.
First, let's move y to one side so we can solve for x.
2x-3y=9
2x-9=3y
y=(2x-9)/3
5x+4y=11
4y=11-5x
y=(11-5x)/4
Now that they both equal the same thing (y), we can set them equal to each other and solve for x. This will give us the x value for the point of intersection of the lines.
(11-5x)/4=(2x-9)/3
3(11-5x)=4(2x-9)
33-15x=8x-36
33+36=8x+15x
69=23x
x=69/23
x=3
Now, we can do the opposite, and solve for x to find the y coordinate.
2x-3y=9
2x=3y+9
x=(3y+9)/2
5x+4y=11
5x=11-4y
x=(11-4y)/5
(3y+9)/2=(11-4y)/5
5(3y+9)=2(11-4y)
15y+45=22-8y
15y+8y=22-45
23y= -23
y= -1
The coordinates for the point of intersection of the two lines is (3, -1).
Answer:
y=2/3x−4
Step-by-step explanation:
If by "long leg lengths" you mean the hypotenuse then the area is 116 sq. units. If you mean the bases of the triangles then the area is 170 sq. units.
If the length of 12 is the hypotenuse, we first must find the base of the triangles using the Pythagorean theorem:
10² + b² = 12²
100 + b² = 144
b² = 44
b = √44 = 6.6
This means we have two triangles and a rectangle. The area of the rectangle is 5(10) = 50 sq. units. The area of each triangle is 1/2(6.6)(10) = 33. Adding all 3 together we have:
50+33+33 = 116 sq. units.
If the 12 is the base, then we have the rectangle with the area of 5(10) = 50 and two triangles each with an area of 1/2(12)(10) = 60:
50+60+60 = 170 sq. units.
It is either A or B. I am not really sure.
