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3 years ago

How can I find x? Click for picture.

Mathematics

1 answer:

romanna [79]3 years ago

6 0

Answer: You add up all the terms given of EACH angle, and make this an addition problem that is "equal to 180" ; since all triangles have three angles that add up to 180 degrees. Then you solve for "x".
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(as follows) ; to get: " x = 29 ⅔ ° " .
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3x + 4 + 2x + x + 8 = 180 ;
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Combine the "like terms" ;

3x + 2x + x = 6x ;

4 + 8 = 12

So, we have: 6x + 12 = 180 ;

Now, subtract "12" from each side of the equation:
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6x + 12 − 12 = 180 − 12 ;
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to get:

6x = 178 ;

Now, divide EACH SIDE of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ;
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6x / 6 = 178 / 6 ;

to get: x = 178/6 = 29 ⅔ ° .
______________________________________
x = 29 ⅔ ° .
______________________________________

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See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$