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2 years ago

1. y - 5 = -2(x - 3) in standard form (pls explain)

Mathematics

1 answer:

Ann [662]2 years ago

6 0

Answer:

The equation is standard form is 2x + y = 11.

The equation is standard form is 2x - 3y = 18

Step-by-step explanation:

Standard form: Ax + By = C

You need to solve the equation so that it is in that form.

y - 5 = -2(x - 3)

y - 5 = -2(x) - 2(- 3) Distributive property

y - 5 = -2x + 6 Simply . remember -2(-3) = 6

y - 5 + 2x = 2x + 6 Adding 2x to both sides of the equation

2x + y - 5 = 6 Now the x-term and the y-term are on one side

2x + y - 5 + 5 = +5 Adding +5 to both sides of the equation

2x + y = 11

Now the equation is standard form is 2x + y = 11

2. y + 4 = 2/3 (x - 3)

y + 4 = 2/3(x) + 2/3(- 3) Distributive property

y + 4 = 2/3x - 2 Simply : 2/3(-3) = -6/3 = -2

y + 4 - 2/3x = -2/3x Adding -2/3x to both sides

y + 4 -2/3x = -2

-4 = -4 Adding -4 to both sides

y - 2/3x = - 6

-y + 2/3x = 6 x- term is positive so multiply by -1 will

2/3x - y = 6 change the sign in the equation

3(2/3x) -3(y) = 3(6) You do not want a fraction in the x-term so

multiply by 3. Each term by 3.

2x - 3y = 18 3(2/3) = 6/3 = 2

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

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$\large \mathbb{PROBLEM:}$

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

$\large \mathbb{SOLUTION:}$

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