All categories

2 years ago

I Need Help With This Problem Please :)

Mathematics

1 answer:

san4es73 [151]2 years ago

7 0

Answer:

Output = input + 2

Step-by-step explanation:

Simple, take some values to figure it out.

When input is 0, output is 2

When input is 1, output is 3

When input is 2, output is 4

When input is 3, output is 5

There is one commonality between them:

The output is simply input + 2

You might be interested in

Which of the following methods would be the easiest to use to solve x2 + 5x – 6 = 0?

EastWind [94]

For me personally, the easiest way to do this is by isolating the x² term, and finding the square root of both sides. The hardest way (well actually, the longest way) would be to use the quadratic formula. It just complicates things unnecessarily.

5 0

2 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

4 0

2 years ago

a major metroplitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. they aske

Arisa [49]

Answer:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.