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2 years ago

Perform the indicated operation. (2x + y)(3x2 + y)

Mathematics

1 answer:

UkoKoshka [18]2 years ago

5 0

Answer:

Step-by-step explanation:

(2x+y)(3x²+y)=2x(3x²+y)+y(3x²+y)=6x³+2xy+3x²y+y²=6x³+3x²y+2xy+y²

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complete the concept map by naming the sampling that you have learned in this module then briefly discuss each

Dmitry [639]

In systematic sampling, individuals are selected at regular intervals from the sampling frame.

<h3>How to explain the sampling?</h3>

Stratified sampling- In this method, the population is first divided into subgroups (or strata) who all share a similar characteristic

<u>Clustered sampling.</u>- a clustered sample, subgroups of the population are used as the sampling unit, rather than individuals. The population is divided into subgroups, known as clusters, which are randomly selected to be included in the study

<u>Simple random sampling. </u>-In this case each individual is chosen entirely by chance and each member of the population has an equal chance, or probability, of being selected.

Quota sampling--This method of sampling is often used by market researchers. Interviewers are given a quota of subjects of a specified type to attempt to recruit.

Learn more about sampling on:

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6 0

1 year ago

Help me please !!!!!

zysi [14]

Nkoooooooooooooooooooooooo

5 0

2 years ago

. -3(2x + 1) = 2(-3x - 1)

Marina CMI [18]

No solution to this problem

4 0

2 years ago

A mail carrier at the U.S. Postal Service works on Sunday delivering high-priority letters and packages. She can deliver 14 ite

Kobotan [32]

Answer:

$4.257

Step-by-step explanation:

Given:

Items delivered per hour = 14

Fixed hourly costs = $47

variable cost = $0.90

Now,

at breakeven point, the profit = 0

Total revenue = total cost

let the charge at breakeven point be 'x'

thus,

x × 14 = $47 + ($0.90 × 14)

x × 14 = $47 + $12.6

14x = $59.6

x = $4.257

Hence, the breakeven charges are $4.257

7 0

2 years ago

Let f be the function defined by f(x) = e^(x) cos x.

Pavel [41]

(a)

The average rate of change of f on the interval 0 ≤ x ≤ π is

$\displaystyle
f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}$

____________

(b)

$f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\
f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\
f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}$

The slope of the tangent line is $e^{3\pi/2}$ .

____________

(c)

The absolute minimum value of f occurs at a critical point where f'(x) = 0 or at endpoints.

Solving f'(x) = 0

$f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\
0 = e^x \big( \cos(x) - \sin(x)\big)$

Use zero factor property to solve.

$e^x \ \textgreater \ 0\forall x \in \mathbb{R}$ so that factor will not generate solutions.
Set cos(x) - sin(x) = 0

$\cos (x) - \sin (x) = 0 \\
\cos(x) = \sin(x)$

cos(x) = 0 when x = π/2, 3π/2, but x = π/2. 3π/2 are not solutions to the equation. Therefore, we are justified in dividing both sides by cos(x) to make tan(x):

$\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\
x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi]$

We check the values of f at the end points and these two critical numbers.

$f(0) = e^1 \cos(0) = 1$

$\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4} \frac{\sqrt{2}}{2}$

$\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4} \frac{-\sqrt{2}}{2} = -e^{\pi/4} \frac{\sqrt{2}}{2}$

$f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}$

There is only one negative number.
The absolute minimum value of f <span>on the interval 0 ≤ x ≤ 2π is
$-e^{5\pi/4} \sqrt{2}/2$

____________

(d)

The function f is a continuous function as it is a product of two continuous functions. Therefore, $\lim_{x \to \pi/2} f(x) = f(\pi/2) = e^{\pi/2} \cos(\pi/2) = 0$

g is a differentiable function; therefore, it is a continuous function, which tells us $\lim_{x \to \pi/2} g(x) = g(\pi/2) = 0$ .

When we observe the limit $\displaystyle \lim_{x \to \pi/2} \frac{f(x)}{g(x)}$ , the numerator and denominator both approach zero. Thus we use L'Hospital's rule to evaluate the limit.

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{f'(\pi/2)}{g'(\pi/2)}$

$f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\
g'(\pi/2) = 2$

thus

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}$ </span>

3 0

2 years ago