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charle [14.2K]
3 years ago
15

Which of the following is a geometric series? 40 + 50 + 60 + 70 40 + 42 + 44 + 46 40 + 80 + 40 + 120 40 + 20 + 10 + 5

Mathematics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

40 + 20 + 10 + 5 is a geometric series

Step-by-step explanation:

* Lets revise the geometric series

- There is a constant ratio between each two consecutive numbers

- Ex:

# 5  ,  10  ,  20  ,  40  ,  80  ,  ………………………. (×2)

# 5000  ,  1000  ,  200  ,  40  ,  …………………………(÷5)

* General term (nth term) of a Geometric Series:

∵ U1 = a  ,  U2  = ar  ,  U3  = ar2  ,  U4 = ar3  ,  U5 = ar4

∴ Un = ar^n-1, where a is the first term , r is the constant ratio

  between each two consecutive terms , and n is the position of

  the term in the series

* Now lets check the answers to find the correct one

# 40 + 50 + 60 + 70

∵ 50/40 = 5/4

∵ 60/50 = 6/5

∵ 5/4 ≠ 6/5

- No common ratio between each two consecutive terms

∴ Not geometric series

# 40 + 42 + 44 + 46

∵ 42/40 = 21/20

∵ 44/42 = 22/21

∵ 21/20 ≠ 22/21

- No common ratio between each two consecutive terms

∴ Not geometric series

# 40 + 80 + 40 + 120

∵ 80/40 = 2

∵ 40/80 = 1/2

∵ 2 ≠ 1/2

- No common ratio between each two consecutive terms

∴ Not geometric series

# 40 + 20 + 10 + 5

∵ 20/40 = 1/2

∵ 10/20 = 1/2

∵ 5/10 = 1/2

- There is a common ratio between each two consecutive terms

∴ 40 + 20 + 10 + 5 is a geometric series

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Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.
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The area of the part of the plane 3x 2y z = 6 that lies in the first octant  is  mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is  mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)

The partial derivatives of a function are f x and f y.

\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\

&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}

In conclusion,  the area is

A=3 √4 units ^2

Read more about the plane

brainly.com/question/1962726

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