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3 years ago

Solve the given equation

Mathematics

1 answer:

AlexFokin [52]3 years ago

4 0

Eliminate the fractions by multiplying by least common multiple.
4,9,12 all go into 36 evenly. Multiply entire equation by 36.

36/4 = 9
36/9 = 4
36/12 = 3

New equation looks like:
$9(w-1) +4(w+1) = -3(w+1) \\ \\ 9w -9 +4w +4 = -3w -3 \\ \\ 9w +4w+3w = -3 +9 -4 \\ \\ 16w = 2 \\ \\ w = \frac{2}{16} = \frac{1}{8}$

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Determine the approximate circumference of a circle with diameter 18 units.

Sphinxa [80]

Answer: 56.549 or 56.5

Explanation:

C= (pie) 3.14 * d = 2 * 3.14 * r

D= 18

C = 3.14 * 18 = 18 * 3.14

C = 56.549 or 56.5

8 0

3 years ago

Determine f^-1(2) given the following: f(1 = -3f(2) = -2f(-1) = 2f(3) = -1

lana [24]

We assume that the function is as follows:

$f^{-1_{}}(2)=?$

Then, we have that:

$f^{-1}(x)=\frac{1}{f(x)}$

Thus, if we have that:

$f(2)=-2\Rightarrow f^{-1^{}_{}}(x)=\frac{1}{f(x)}\Rightarrow f^{-1}(2)=\frac{1}{f(2)}=\frac{1}{-2}\Rightarrow f^{-1}(2)=-\frac{1}{2}$

Therefore, the value for the function is f^-1(2) = -1/2 or

$f^{-1}(2)=-\frac{1}{2}$

8 0

1 year ago

William is going on vacation and taking his dog Pickles along. Pickles eats 3 cups of Hungry

prohojiy [21]

Answer:

5 days

Step-by-step explanation:

He eats 3 cups a day and its a 15 cup bag,

15 divided by 3 equals 5.

So he can feed Pickles for 5 days from the bag.

8 0

2 years ago

Write in slope-intercept form an equation of the line that passes through the given points.

nlexa [21]

Answer: y = 1/ 8 x − 1

8 0

3 years ago

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

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

Step-by-step explanation:

We are given the following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

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

Unit: Differentiation

5 0

2 years ago