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:
y=x-5
Step-by-step explanation:
slope of line s=-1= m1
lines are perpendicular m1m2=-1 m2=1
equations of t=y-y1=m2(x-x1)
y-2=x+3
y=x+5
Answer:
45,78%
Step-by-step explanation:
42 + 67 = 109
109 x 42 / 100 = 45,78 %
Answer:
The value of 
is 
.
Step-by-step explanation:
The given equation is
We need to find the value of .
Differentiate with respect to t.
![[\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cfrac%7Bd%7D%7Bdx%7Dx%5En%3Dnx%5E%7Bn-1%7D%2C%5Cfrac%7Bd%7D%7Bdx%7DC%3D0%5D)
It is given that y=2 and dy/dt=1, substitute these values in the above equation.
Divide both sides by 4x³.
Therefore the value of is .
