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3 years ago

I need help, Please and Thank you

Mathematics

1 answer:

mr Goodwill [35]3 years ago

8 0

Graph 1 the domain and range are both all real numbers.
Graph 2 the domain is [-1,5] and range is [-1,2]

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In the function y=5x^2-2, what effect does the number 5 have on the graph, as compared to the graph of y=x^2?

allochka39001 [22]

General Idea:

The Rules for Transformations of Functions are given below:

If f(x) is the original function, a > 0 and c > 0; Then

$f(x)+k \; \; shift \; f(x) \; UPWARD \; k \; units\\\\f(x)-k \; \; shift \; f(x) \; DOWNWARD \; k \; units\\\\f(x+h) \; \; shift \; f(x) \; LEFT \; h \; units\\\\f(x-h) \; \; shift \; f(x) \; RIGHT \; h \; units\\\\-f(x) \; \; REFLECT \; f(x) \; in \; the \; x-axis\\\\f(-x) \; REFLECT \; f(x) \; in \; the \; y-axis\\\\a \cdot f(x), \; a>1 \; STRETCH \; f(x) \; vertically \; by \; a \; factor \; of \; a\\\\a \cdot f(x), \; 0$

$f(ax), \; a>1 \; SHRINK \; f(x) \; horizontally \; by \; a \; factor \; of \;\frac{1}{a} .\\\\f(ax), \; 0$

Applying the concept:

In the function y=5x^2-2, the effect that the number 5 have on the graph, as compared to the graph of y=x^2 is given below:

C.it stretches the graph vertically by a factor of 5

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3 years ago

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Priya has five pencils each X inches in length when she lines of pencils and and they measure 34.5 in. Select all the equations

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Answer:

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Expressions is equivalent to a(b-c)-b(a+c)-c(a-b)?

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Which system is the solution of the graph? HELPPPP

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3 years ago

What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)

zzz [600]

Answer:

B) 4√2

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
Integration Constant C

Arc Length Formula [Parametric]: $\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx$

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: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

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