Find the sum of a finite geometric series.
2 answers:
Answer with explanation:
⇒Side of largest equilateral triangle in which three equilateral triangles are inscribed = 16 inches
Perimeter of a triangle = Sum of three sides of triangle
Perimeter of equilateral triangle having side length 16 inches = 16 +16+16=48 inches
⇒→Second equilateral triangle which is inscribed in this equilateral triangle having side length half of that equilateral triangle in which it is inscribed
inches
Perimeter of equilateral triangle having side length 8 inches = 8 +8+8=24 inches
⇒→Third equilateral triangle which is inscribed in this equilateral triangle having side length half of that equilateral triangle in which it is inscribed
inches
Perimeter of equilateral triangle having side length 4 inches =4+4+4=12 inches
⇒→Fourth equilateral triangle which is inscribed in this equilateral triangle having side length half of that equilateral triangle in which it is inscribed
inches
Perimeter of equilateral triangle having side length 2 inches =2+2+2=6 inches
→≡Total Perimeter of all four Equilateral Triangle
=48 +24+12+6
= 90 inches
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Answer:
The Line integral is π/2.
Step-by-step explanation:
We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:
we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence
(Remember that c is treated like a constant just for the x-variable).
This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.
Where, again, the constant of integration depends on Z.
As a result,
if we derivate f over z, we obtain
That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)
The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.