Decide whether the relation defines a function.

Mathematics

1 answer:

diamong [38]3 years ago

3 0

Answer:

Step-by-step explanation:

Lets first solve the relation for "y". Shown below:

$3x=9-4y\\4y=-3x+9\\y=-\frac{3}{4}x+\frac{9}{4}$

This is clearly an equation of a line in the form $y=mx+c$

It will be a negatively sloped line, when graphed.

We can use vertical line test to determine function or relation. If a vertical line passes through only 1 point in the graph, we say that it is a function.

Think of a negatively sloped line. No matter where you draw a vertical line, it will only cut the line at 1 point.

So, this is definitely an equation of a function

Correct answer is A

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General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

Integrals
Definite Integrals
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Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.$

Interval [0, π]

Step 2: Find Arc Length

[Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]: $\displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.$
Substitute in variables [Arc Length Formula - Parametric]: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx$
[Integrand] Simplify: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx$
[Integral] Evaluate: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$