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2 years ago

The equation, 3x2 6x 3y2 7y 4 = 0, represents a conic. The conic represented by the equation is a/an because.

1 answer:

Taya2010 [7]2 years ago

3 0

The given equation is an equation of a circle because it looks similar to the general equation of a circle.

<h3>What is the general equation of a circle?</h3>

The general equation of a circle is:

x² + y² +2gx + 2fy + c = 0...........eq1

Where (-g, -f) is the center of the circle.

c is a constant

Given equation is

3x² +6x + 3y²+7y+4 = 0

Let us take 3 as common

x² + 2x+y²+7/3y + 4/3 = 0

Let us rearrange the equation

x² + y²+2x +7/3y + 4/3 = 0...........eq2

Now look at eq1 and eq2

We got that eq2 is having a similar pattern as eq1.

Therefore, The given equation is an equation of a circle.

To get more about circle visit:

brainly.com/question/1506955

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Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A What is the value of the A^2?

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$

$\boxed{ \tt \red{C}arry \: \red{ O}n \: \red{L}earning} \: \underline{\tt{5/13/22}}$

4 0

2 years ago

Find cos(β) in the triangle.

svetlana [45]

90°+b+b =180

90°+ 2b=180

2b=180-90

2b= 90

2b=90

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b=45°

ave tried I don't know if its correct

8 0

2 years ago

three bells ring at the intervals of 10 minutes and 20 minutes if they all ring together at 10 clock when will they ring togethe

mario62 [17]

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Step-by-step explanation:

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3 years ago

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wolverine [178]

Answer:

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Step-by-step explanation:

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3 years ago

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AveGali [126]

18 11 25 8 4 !!!!!!!!

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3 years ago