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4 years ago

Please help! attechment is posted below will give brainliest rattings thanks etc

Mathematics

1 answer:

victus00 [196]4 years ago

6 0

Answer:

Choice B.

Step-by-step explanation:

In a function, each x-coordinate can only appear in one single point. In a relation, if more than one point have the same x-coordinate, it is not a function.

Apply the vertical line test.

Imagine a vertical line starting at the left of the graph and moving to the right.

If the line in any position intersects more than one point of the relation, then the relation is not a function.

If the line in every position only intersects one point at a time, then the relation is a function.

Start with Choice A.

A vertical line starts at the left and moves right. It intersects point (-4, 3) by itself. Then it intersects point (-2, 1) by itself. When the vertical line gets to x = -1, it intersects two points, (1, -1) and (-1, 4). This means it is not a function.

Choice B.

Any vertical line in any position intersects at most one point at a time. Choice B is a function.

Choice C.

At x = -2 and at x = 2, a vertical line intersects two points. This is not a function.

Choice D.

At x = 0, a vertical line intersects two points. This is not a function.

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Marizza181 [45]

Answer:

Point A(9, 3)

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: $\displaystyle x = 9$