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3 years ago

Solve equation for 5x-14=8x+4

Mathematics

2 answers:

creativ13 [48]3 years ago

6 0

My answer-

Simplifying 5x + -14 = 8x + 4

Reorder the terms: -14 + 5x = 8x + 4 Reorder the terms: -14 + 5x = 4 + 8x

Solving -14 + 5x = 4 + 8x Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right. Add '-8x' to each side of the equation. -14 + 5x + -8x = 4 + 8x + -8x

Combine like terms: 5x + -8x = -3x -14 + -3x = 4 + 8x + -8x

Combine like terms: 8x + -8x = 0 -14 + -3x = 4 + 0 -14 + -3x = 4 Add '14' to

each side of the equation. -14 + 14 + -3x = 4 + 14

Combine like terms: -14 + 14 = 0 0 + -3x = 4 + 14 -3x = 4 + 14

Combine like terms: 4 + 14 = 18 -3x = 18 

Divide each side by '-3'. x = -6

Simplifying x = -6

If you need anything else on brainly let me know :)

IgorC [24]3 years ago

6 0

Remember, you can do anything to an equation as long as you do it to both sides

get 1 of the x on one side

5x-14=8x+4
minus 5x both sides (we want on left side because 8>5 and I like positive stuff)

5x-5x-14=8x-5x+4
0-14=3x+4
-14=3x+4
minus 4 both sides
-14-4=3x+4-4
-18=3x+0
-18=3x
divide both sides by 3
remember, negative/positive=negative
-18/3=(3x)/3
-6=x(3/3)
-6=x(1)
-6=x
x=-6

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Answer:

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The complex roots occur in conjugate pairs so there are 3 roots 2, 3i and -3i.

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2 years ago

Read 2 more answers

Help with part b ! - please see attachment

vfiekz [6]

Answer:

(b) $\displaystyle \int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx = \arctan \big( \sec x \big) + C$

General Formulas and Concepts:
Calculus

Differentiation

Derivatives
Derivative Notation

Integration

Integrals

Integration Method: U-Substitution and U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle \int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u:
$\displaystyle u = \sec x$
[u] Apply Trigonometric Differentiation:
$\displaystyle du = \sec x \tan x \ dx$
[du] Rewrite [U-Solve]:
$\displaystyle dx = \cos x \cot x \ du$

Step 3: Integrate Pt. 2

[Integral] Apply Integration Method [U-Solve]:
$\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \leftarrow \\\end{aligned}$
[Integrand] Simplify:
$\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \leftarrow \\\end{aligned}$
[Integral] Apply Arctrigonemtric Integration:
$\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \leftarrow \\\end{aligned}$
Simplify:
$\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \\& = \arctan u + C \leftarrow \\\end{aligned}$
[u] Back-substitute:
$\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \\& = \arctan u + C \\& = \boxed{ \arctan \big( \sec x \big) + C } \\\end{aligned}$

∴ we used substitution to find the indefinite integral.

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Learn more about integration: brainly.com/question/27746468

Learn more about Calculus: brainly.com/question/27746481

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

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