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3 years ago

The points (3,0) and (6,4) are adjacent vertices of a square, What is the area of the square, in unit squared? HELP PLS

Mathematics

2 answers:

zaharov [31]3 years ago

7 0

Answer: 25

Step-by-step explanation:

Find the distance from (3, 0) to (6,4). That is the side length (s).

Area (A) = s²

$s=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\s^2=(x_2-x_1)^2+(y_2-y_1)^2\\\\s^2=(6-3)^2+(4-0)^2\\\\s^2=3^2+4^2\\\\s^2=9+16\\\\s^2=25\\\\\\A=s^2\\\\\large\boxed{A = 25}$

mestny [16]3 years ago

6 0

Answer:

25 units squared

Step-by-step explanation:

Distance between (3, 0) and (6, 4)

√(3-6)²+(0-4)² = 5

Area = 5 * 5 = 25 units squared

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

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:

Step 1: Define

Identify

$\displaystyle y = \sqrt{x}$

$\displaystyle y' = \frac{1}{6}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = x^{\frac{1}{2}}$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x}}$

Step 3: Solve

Find coordinates of A.

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}$
[Multiplication Property of Equality] Cross-multiply: $\displaystyle \sqrt{x} = 3$
[Equality Property] Square both sides: $\displaystyle x = 9$