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1 year ago

4

Can you explain the process on how to solve this?

1 answer:

oee [108]1 year ago

8 0

Given the expression:

$\frac{x+4}{x}-\frac{3}{2}=-\frac{1}{x}$

We can rewrite it as:

$\frac{x+4}{x}+\frac{1}{x}=\frac{3}{2}$

Solving the expression:

$\begin{gathered} \frac{x+4+1}{x}=\frac{3}{2} \\ \\ \frac{x+5}{x}=\frac{3}{2} \end{gathered}$

Cross multiplication:

2x + 10 = 3x

Solving for x:

3x - 2x = 10

x = 10

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Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

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Answer:

Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16% Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%. Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%. Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%Suppose you have a mean = 12 and a standard deviation = 3. What is the probability that a data member, x, is above 18?

A) 34%

B) 2.5%

C) 5%,

D) 16%

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3 years ago

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erastovalidia [21]

Answer:

See explanation.

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients
Factoring

<u>Algebra II</u>

Polynomial Long Division

<u>Pre-Calculus</u>

Parametrics

<u>Calculus</u>

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

Parametric Differentiation: $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle x = 2t - \frac{1}{t}$

$\displaystyle y = t + \frac{4}{t}$

<u>Step 2: Find Derivative</u>

[<em>x</em>] Differentiate [Basic Power Rule and Quotient Rule]: $\displaystyle \frac{dx}{dt} = 2 + \frac{1}{t^2}$
[<em>y</em>] Differentiate [Basic Power Rule and Quotient Rule]: $\displaystyle \frac{dy}{dt} = 1 - \frac{4}{t^2}$
Substitute in variables [Parametric Derivative]: $\displaystyle \frac{dy}{dx} = \frac{1 - \frac{4}{t^2}}{2 + \frac{1}{t^2}}$
[Parametric Derivative] Simplify: $\displaystyle \frac{dy}{dx} = \frac{t^2 - 4}{2t^2 + 1}$
[Parametric Derivative] Polynomial Long Division: $\displaystyle \frac{dy}{dx} = \frac{1}{2} - \frac{7}{2(2t^2 - 1)}$
[Parametric Derivative] Factor: $\displaystyle \frac{dy}{dx} = \frac{1}{2} \bigg( 1 - \frac{9}{2t^2 + 1} \bigg)$

Here we see that if we increase our values for <em>t</em>, our derivative would get closer and closer to 0.5 but never actually reaching it. Another way to approach it is to take the limit of the derivative as t approaches to infinity. Hence $\displaystyle \frac{dy}{dx} < \frac{1}{2}$ .

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametrics

Book: College Calculus 10e

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