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:
See explanation.
Step-by-step explanation:
We are looking at a geometric distribution.
The probability of selecting a brown peanut is .12 = p
The probability of not selecting a brown peanut is .88 = q
The probability mass function is p(y) = (.88)^(y-1) * (.12)
a) p(7) = (.88)^6 * .12 = .0557
b) p(7 <= y <= 8) = p(7) + p(8)
= .0557 + (.88)^7 * .12 = .1048
c) p(y <= 7) = p(0) + p(1) + ... + p(7)
= .12 + (.88)^1 * .12 + (.88)^2 * .12 + ... + (.88)^6 * .12 = .4713
d) The expect value is 1/p. So, 1/(.12) = 8.33 M&M's
Answer:
192 cm^3/min
Step-by-step explanation:
Differentiating the volume expression, we get ...
dV/dt = 2xh(dx/dt) +x^2(dh/dt)
We are given that ...
x = 4 cm, dx/dt = 2 cm/min, h = 15 cm, dh/dt = -3 cm/min
Putting these values into the formula for volume rate of change, we get ...
dV/dt = 2(4 cm)(15 cm)(2 cm/min) +(4 cm)^2(-3 cm/min)
= 240 cm^3/min -48 cm^3/min
dV/dt = 192 cm^3/min
Answer:
a) 15.52ft
b) 0.47 secs
Step-by-step explanation:
h = -16t^2 + 15t + 12
The maximum height is reached after -b/2a seconds
a = -16 , b = 15, c= 12
-b/2a = -15/2(-16)
= -15/-32
= 0.47 secs
The maximum height = -16(0.47)^2 + 15(0.47) + 12
= -3.5344 + 7.05 + 12
= 15.5156
= 15.52ft
Answer:
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