All categories

2 years ago

The number of members in a labor union is 240, and the number increases by 5% each year. Find the number of members after 4 year

1 answer:

4 0

$~~~~~~ \textit{Union Members Earned Amount}\\\\A=P(1+rt)\qquad\begin{cases}A=\textit{accumulated amount}\\P=\textit{original amount}\dotfill & 240\\r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\t=years\dotfill &4\end{cases}\\\\\\A=240[1+(0.05)(4)]\implies A=240(1.2)\implies A=288$

You might be interested in

Help please!!!!:::_))

sleet_krkn [62]

Answer with Step-by-step explanation:

Independent:Tickets,Shopping list,weight,Hiking,Mushrooms in the bridge,Number of trophies

Dependent:Money,Shopping bas,price of customer's order,snacks,mushroom tarts, shelves on the case

5 0

2 years ago

Find the missing height of the cone. Volume= 118π ft3 Diameter= 2/3

grin007 [14]

Answer:

11 ft

Step-by-step explanation:

The equation for the volume of a cone is:

V = 1/3(3.14)r^2h

V = 414.48

r = 6

Plug in those values and solve for h:

414.48 = 1/3(3.14)(6)^2h

414.48 = 37.68h ; divide both sides by 37.68

11 = h

The missing height is 11ft.

6 0

2 years 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:

Calculus

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 x = 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 x = 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 x = 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

If m<7 is 2x + 30 and m<28 is 3x - 10, find the following:

topjm [15]

Answer:

x = 32

m∠7 = 94

m∠8 = 86

m∠3 = 94

Step-by-step explanation:

(2x + 30) + (3x - 10) = 180

5x + 20 = 180

5x = 160

x = 32

m∠7 = 2(32) + 30 = 94

m∠8 = 3(32) - 10 = 86

or

m∠8 = 180 - 94 = 86

m∠3 ≈ m∠7 (corresponding angles)

3 0

3 years ago

ANSWER ASAP PLEASE

inna [77]

Same as the dude that did the same thing

6 0

2 years ago

Read 2 more answers