6i/ (1+i)
multiply by the complex conjugate (1-i)/(1-i)
6i/(1+i) * (1-i)/(1-i)
6i* (1-i) = 6i - 6i^2 = 6i - 6(-1) = 6i +6
(1+i)*(1-i)= 1-i +i -i^2 = 1 -i+i -(-1) = 1+1=2
(6+6i)/2
3+3i
Answer: 3+3i
As of 12:04 EST U.S.
$1=<span>112.624847Yen
So:
100USD(112.624847Y/1USD)=11262.62 Yen</span>
Answer:
10
Step-by-step explanation:
If there is a direct relation between two variables x and y then it can be represented as
y = kx ,
where y is dependent variable
x is independent variable
k is constant of variation
_____________________________
First condition
y = 400
x = r
using y = kx then relationship will be
400 = kr
finding k here
k = 400/r
Second condition
y = r
x = 4
using y = kx then relationship will be
r = 4k
finding k
k = r/4
since in both condition equation is same
thus, value of k will also be same
thus,
400/r = r/4
=> 400*4 = r*r
=> 1600 = r^2

Thus, 40 is the value of r
k = r/4 = 40/4 = 10
Thus, constant of variation is 10 which is correct choice.
To cross validate
k = 400/r = 400/40 = 10
Answer:
455 or 680, depending
Step-by-step explanation:
If we assume the three choices are different, then there are ...
15C3 = 15·14·13/(3·2·1) = 35·13 = 455
ways to make the pizza.
___
If two or three of the topping choices can be the same, then there are an additional ...
2(15C2) +15C1 = 2·105 +15 = 225
ways to make the pizza, for a total of ...
455 + 225 = 680
different types of pizza.
__
There is a factor of 2 attached to the number of choices of 2 toppings, because you can have double anchovies and tomato, or double tomato and anchovies, for example, when your choice of two toppings is anchovies and tomato.
_____
nCk = n!/(k!(n-k)!)
The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin.
<span>Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. </span>
<span>If F is the mean of f over the region R then F ∫ (R)dV = ∫ (R)fdV </span>
<span>∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz </span>
<span>= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 </span>
<span>∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz </span>
<span>= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ </span>
<span>= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ </span>
<span>= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 </span>
<span>∴ F = 10935π/8 ÷ 81π/2 = 135/4</span>