twelve rolls of commercial-grade paper towels come in one case costing $23.58 What is the the cost per roll of paper towels

Suppose you earn $6 per hour at your part time job what will your new rate be after a 2.5% raise

ivanzaharov [21]

2.5% means 0.025 .
After an increase of 2.5%, the new rate is (1.025) x (the old rate).

1.025 x $6.00 = $6.15 per hour.

If you work a full 40 hours in some week, you will be blessed with
a full $6 more for the week of toil than you would have collected
at the old rate.

You were not overpaid before the raise, and you must not allow
your boss to act as if you should kiss his ring or prostrate yourself
before him in return for 2.5% . It was not too long ago that 2.5% was
tantamount to being shown the way to the door, and a tacit suggestion
that you consider following in the way of the arrow.

5 0

3 years ago

What is the factored form of 30x2 - 25x -30?

Setler79 [48]

Answer:

Step-by-step explanation:

Here are some steps to help you

Step 1

30x2 - 25x -30 Factoring

Step 2

30x2 - 25x -30 factor out the equation

5(3x+2)(2x-3)

Step 3

5(3x+2)(2x-3) Check by simplifying

When you check, you will get the question back

Answer

5(3x+2)(2x-3)

Hope this helps

6 0

3 years ago

A and B are independent events.

kiruha [24]

Answer:

Not enough information

5 0

3 years ago

What value of X is always

valentina_108 [34]

Answer:

Answer B

Step-by-step explanation:

Because (IQR) by 1.5 (a constant used to discern outliers). Add 1.5 x (IQR) to the third quartile. Any number greater than this is a suspected outlier. Subtract 1.5 x (IQR) from the first quartile.

5 0

3 years ago

Help with num 3 please. thanks

Alja [10]

Answer:

a) $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1$

b) $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = \frac{\pi}{2}} = -1$

General Formulas and Concepts:

Pre-Calculus

Unit Circle

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Trigonometric Differentiation

Logarithmic Differentiation

Step-by-step explanation:

a)

Step 1: Define

Identify

$\displaystyle y = ln \bigg( \frac{1 - x}{\sqrt{1 + x^2}} \bigg)$

Step 2: Differentiate

Logarithmic Differentiation [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 - x}{\sqrt{1 + x^2}}} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]$
Quotient Rule: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{(1 - x)'\sqrt{1 + x^2} - (1 - x)(\sqrt{1 + x^2})'}{(\sqrt{1 + x^2})^2}$
Basic Power Rule [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{-\sqrt{1 + x^2} - (1 - x)(\frac{x}{\sqrt{x^2 + 1}})}{(\sqrt{1 + x^2})^2}$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \bigg( \frac{x(x - 1)}{(x^2 + 1)^\bigg{\frac{3}{2}}} - \frac{1}{\sqrt{x^2 + 1}} \bigg)$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{x + 1}{(x - 1)(x^2 + 1)}$

Step 3: Find

Substitute in x = 0 [Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = 0} = \frac{0 + 1}{(0 - 1)(0^2 + 1)}$
Evaluate: $\displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1$

b)

Step 1: Define

Identify

$\displaystyle y = ln \bigg( \frac{1 + sinx}{1 - cosx} \bigg)$

Step 2: Differentiate

Logarithmic Differentiation [Chain Rule]: $\displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 + sinx}{1 - cosx}} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]$
Quotient Rule: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{(1 + sinx)'(1 - cosx) - (1 + sinx)(1 - cosx)'}{(1 - cosx)^2}$
Trigonometric Differentiation: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{cos(x)(1 - cosx) - sin(x)(1 + sinx)}{(1 - cosx)^2}$
Simplify: $\displaystyle \frac{dy}{dx} = \frac{-[cos(x) - sin(x) - 1]}{[sin(x) + 1][cos(x) - 1]}$

Step 3: Find

Substitute in x = π/2 [Derivative]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = \frac{-[cos(\frac{\pi}{2}) - sin(\frac{\pi}{2}) - 1]}{[sin(\frac{\pi}{2}) + 1][cos(\frac{\pi}{2}) - 1]}$
Evaluate [Unit Circle]: $\displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = -1$

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

Unit: Differentiation

Book: College Calculus 10e

4 0

3 years ago