A total of 250 people were surveyed about whether or not each person carries a cell phone.

Solve 2x+4 so that it has one solution

natima [27]

Answer:

-2

Step-by-step explanation:

We can set it equal to zero

2x+4 = 0

2x =-4

x = -2

7 0

3 years ago

Read 2 more answers

Isaac is looking through binoculars on a whale watching trip when he notices a sea otter in the distance. If he is 20 feet above

sveta [45]

Tan of an angle = perpendicular / base
Here, perpendicular is the side of the triangle opposite the angle. 
And base is the arm of the angle other the hypotenuse. 

If the side of the triangle to be found (marked as ? in the figure) is x, then 

perpendicular = 20 feet 
base = x 

so, tan 30 = 20/x 
so, x = 20/(tan 30 degrees) = 34.64 feet 
= 35 feet to the nearest foot

6 0

3 years ago

Read 2 more answers

Emma is moving and must rent a truck. In addition to an initial fee, the rental company charges a fee of $2.50 per mile driven.

Akimi4 [234]

Answer:

C=2.50m+20

Step-by-step explanation:

8 0

3 years ago

Find the derivative: y={ (3x+1)cos(2x) } / e^2x

DochEvi [55]

Answer:

$\displaystyle y' = \frac{3cos(2x) -2(3x + 1)[sin(2x) + cos(2x)]}{e^{2x}}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring
Exponential Rule [Dividing]: $\displaystyle \frac{b^m}{b^n} = b^{m - n}$
Exponential Rule [Powering]: $\displaystyle (b^m)^n = b^{m \cdot n}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

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

Trig Derivative: $\displaystyle \frac{d}{dx}[cos(u)] = -u'sin(u)$

eˣ Derivative: $\displaystyle \frac{d}{dx}[e^u] = u'e^u$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \frac{(3x + 1)cos(2x)}{e^{2x}}$

Step 2: Differentiate

[Derivative] Quotient Rule: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - \frac{d}{dx}[e^{2x}](3x + 1)cos(2x)}{(e^{2x})^2}$
[Derivative] [Fraction - Numerator] eˣ derivative: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{(e^{2x})^2}$
[Derivative] [Fraction - Denominator] Exponential Rule - Powering: $\displaystyle y' = \frac{\frac{d}{dx}[(3x + 1)cos(2x)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] Product Rule: $\displaystyle y' = \frac{[\frac{d}{dx}[3x + 1]cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] [Brackets] Basic Power Rule: $\displaystyle y' = \frac{[(1 \cdot 3x^{1 - 1})cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] [Brackets] (Parenthesis) Simplify: $\displaystyle y' = \frac{[3cos(2x) + \frac{d}{dx}[cos(2x)](3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] [Brackets] Trig derivative: $\displaystyle y' = \frac{[3cos(2x) -2sin(2x)(3x + 1)]e^{2x} - 2e^{2x}(3x + 1)cos(2x)}{e^{4x}}$
[Derivative] [Fraction - Numerator] Factor: $\displaystyle y' = \frac{e^{2x}[(3cos(2x) -2sin(2x)(3x + 1)) - 2(3x + 1)cos(2x)]}{e^{4x}}$
[Derivative] [Fraction] Simplify [Exponential Rule - Dividing]: $\displaystyle y' = \frac{3cos(2x) -2sin(2x)(3x + 1) - 2(3x + 1)cos(2x)}{e^{2x}}$
[Derivative] [Fraction - Numerator] Factor: $\displaystyle y' = \frac{3cos(2x) -2(3x + 1)[sin(2x) + cos(2x)]}{e^{2x}}$

Topic: AP Calculus AB/BC

Unit: Derivatives

Book: College Calculus 10e

6 0

3 years ago

Let U = {q, r, s, t, u, v, w, x, y, z}

Sliva [168]

The Set A’∪ B = { r, t, v, x, z, q, s, y, z }

We have the following sets -

U = {q, r, s, t, u, v, w, x, y, z}

A = {q, s, u, w, y}

B = {q, s, y, z}

C = {v, w, x, y, z}

We have to determine the set represented by - A' ∪ B.

A set is a collection of elements or numbers or objects, represented within the curly brackets { }. For example: {1,2,3,4} is a set of numbers.

According to question, we have -

U = {q, r, s, t, u, v, w, x, y, z} (Universal Set)

A = {q, s, u, w, y}

B = {q, s, y, z}

C = {v, w, x, y, z}

Consider Set A = {q, s, u, w, y}

A' = U - A = {q, r, s, t, u, v, w, x, y, z} - {q, s, u, w, y} = { r, t, v, x, z }

and Set B = {q, s, y, z}

Now -

A' ∪ B = { r, t, v, x, z } ∪ {q, s, y, z} = { r, t, v, x, z, q, s, y, z }

Hence, the Set A’∪ B = { r, t, v, x, z, q, s, y, z }

To solve more questions on Set theory, visit the link below -

brainly.com/question/13042571

#SPJ1

6 0

1 year ago