find the length and width of a rectangle that has an area of 968 square feet and whose perimeter is a minimum

find the three consecutive odd integers such that the sum of twice the first and three times the second is 67 more than twice th

Elenna [48]

Answer:

23, 25, 27

Step-by-step explanation:

It can be convenient to work "consecutive integer" problems by using a variable to represent the middle one.

__

<h3>setup</h3>

Let x represent the middle integer. Then (x-2) is the first of the three odd integers, and (x+2) is the third. The given relation is ...

2(x -2) +3x = 67 +2(x +2)

<h3>solution</h3>

5x -4 = 2x +71 . . . . . simplify

3x = 75 . . . . . . . . add 4-2x

x = 25 . . . . . . . divide by 3

The three integers are 23, 25, 27.

4 0

1 year ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

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:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in <em>x</em> = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in <em>x</em> = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

2 years ago

Karen could lift 18 pounds with a

Grace [21]

Answer:

14 more pounds

Step-by-step explanation:

32-18

5 0

3 years ago

Read 2 more answers

Mike can be paid in one of two ways based on the amount of merchandise he sells: Plan A: A salary of $1,000.00 per month, plus a

Serggg [28]

Answer:

X > 21000

Step-by-step explanation:

Given the following :

Payment plans :

PLAN A:

salary = $1000 per month

Commision = 10% of sales

PLAN B:

salary = $1300 per month

Commision = 15% of sales in excess of $9,000

Hence, for plan B; 15% is paid after deducting $9000 from total sales

For what amount of monthly sales is plan B better than plan A if we can assume that Mike's sales are always more than $9,000.00?

That is ;

Plan B > plan A

Let total sales = x

Plan A:

$1,000 + 0.1x

Plan B:

$1,300 + 0.15(x - 9000)

1300 + 0.15(x - 9000) > 1000 + 0.1x

1300 + 0.15x - 1350 > 1000 + 0.1x

0.15x - 50 > 1000 + 0.1x

0.15x - 0.1x > 1000 + 50

0.05x > 1050

x > 1050/0.05

x > 21000

6 0

2 years ago

Find the mistake in this sentence:

wlad13 [49]

The mistake is “there are”

4 0

1 year ago