Diagonal of a Rhombus are perpendicular & intersects in their middle point:
Assume the diagonals intersects in H
A(0,-8), B(1,-0), C(8,-4) & D(x, y) are the vertices of the rhombus and we have to calculate D(x, y)
Consider the diagonal AC. Find the coordinate (x₁, y₁) H, the middle of AC
Coordinate (x₁, y₁) of H, middle of A(0,-8), C(8,-4)
x₁ (0+8)/2 & y₁=(-8-4)/2 ==> H(4, -6)
Now let's calculate again the coordinate of H, middle of the diagonal BD
B(1,-0), D(x, y)
x value = (1+x)/2 & y value=(y+0)/2 ==> x= (1+x)/2 & y=y/2
(1+x)/2 & y/2 are the coordinate of the center H, already calculated, then:
H(4, -6) = [(1+x)/2 , y/2]==>(1+x)/2 =4 ==> x=7 & y/2 = -6 ==> y= -12
Hence the coordinates of the 4th vertex D(7, -12)
Answer:
f(x) = 2x + 22, The slope is 2/1
Step-by-step explanation:
Start by combining common terms.
For example, you can't combine 2x and 5, because 2 is being multiplied by a variable.
So, take the x value: 2x
And the numerical values: 22
And combine them: f(x) = 2x + 22
This means, that you start at 22 on the y-axis and slope the line by 2/1. That means that, on a graph, the line goes up 2 points, then one point to the right.
Answer:
A
Step-by-step explanation:
RT=12
TU=5
RU=12+5=17
RS=17+7=24
SU=√(RS²-RU²)=√(24²-17²)=√(24-17)(24+17)=√(7×41)=√287
area=1/2×12×√287=6×√287≈101.6 unit²
