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3 years ago

Given the Arithmetic sequence A1,A2,A3,A4 45, 51, 57, 63 What is the value of A34?

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

8 0

Answer:

Value of A(34) = 219

Step-by-step explanation:

A1, A2,A3,A4,45,51,57,63 are in Arithmetic sequence

Common difference in this Arithmetic sequence = 63 - 57 = 06

so, A4 = 45 - 6 = 39

A3 = 39 - 6 = 33

A2 = 33 - 6 = 27

A1 = 27 - 6 = 21

First term of Arithmetic sequence = 21

Common difference = 06

Using general term of Arithmetic sequence,

A(n) = First term + (n - 1) common difference

A(34) = 21 + (34 - 1) 6

A(34) = 21 + 33 × 6

A(34) = 21 + 198

A(34) = 219

Value of A(34) = 219

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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:

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

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