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pashok25 [27]
2 years ago
10

Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What

should she use as her common denominator? 6 9 12 18
Mathematics
2 answers:
White raven [17]2 years ago
8 0

Answer:

Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18

Step-by-step explanation:

Ghella [55]2 years ago
4 0

Answer:

18

Step-by-step explanation:

You might be interested in
Find the derivative of following function.
Aleks04 [339]

Answer:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x}  + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}

General Formulas and Concepts:
<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
\displaystyle (cu)' = cu'

Derivative Property [Addition/Subtraction]:
\displaystyle (u + v)' = u' + v'

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
\displaystyle (uv)' = u'v + uv'

Derivative Rule [Quotient Rule]:
\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}

Derivative Rule [Chain Rule]:
\displaystyle [u(v)]' = u'(v)v'

Step-by-step explanation:

*Note:

Since the answering box is <em>way</em> too small for this problem, there will be limited explanation.

<u>Step 1: Define</u>

<em>Identify.</em>

\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}

<u>Step 2: Differentiate</u>

We can differentiate this function with the use of the given <em>derivative rules and properties</em>.

Applying Product Rule:

\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '

Differentiating the first portion using Quotient Rule:

\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}

Apply Derivative Rules and Properties, namely the Chain Rule:

\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}

Differentiating the second portion using Quotient Rule again:

\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Apply Derivative Rules and Properties, namely the Chain Rule again:
\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Substitute in derivatives:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Simplify:

\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}

We can rewrite the differential by factoring and common mathematical properties to obtain our final answer:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x}  + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}

∴ we have found our derivative.

---

Learn more about derivatives: brainly.com/question/26836290

Learn more about calculus: brainly.com/question/23558817

---

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

Unit: Differentiation

8 0
2 years ago
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How do I write 3x3x3x3 in Standard Form
wlad13 [49]

Answer:

3 to the fourth power, which equals 81.

Step-by-step explanation:

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Step-by-step explanation:

corvet corvet

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Write an equation for each figure side
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3 years ago
The sum of 5 consecutive odd integers is 2,555. What is the sum of the first, third, and last integer?
Lilit [14]

Answer:

The sum of the first, third, and last integer is 1533.

Step-by-step explanation:

<u><em>Step 1: You need to create a equation first:</em></u>

n + n + 2 + n + 4 + n + 6 + n + 8 = 2555

<u><em>How did I make the equation?</em></u>

If n is odd number, n + 2 must be the consecutive odd number

5 , 5 + 2 , 5 + 4 , 5 + 6 , 5 + 8

5, 7, 9, 11, 13  ALL CONSECUTIVE ODD NUMBERS

<u><em>Step 2: Simplifying the equation</em></u>

n + n + 2 + n + 4 + n + 6 + n + 8 = 2555

5n + 20 = 2555

<u><em>Step 3: Solving the equation, with explanation</em></u>

5n + 20 = 2555       Subtract 20 on both sides

        5n = 2535        Isolate the variable by dividing 5 on both sides

          n = 507

<u><em></em></u>

<u><em>Step 4: Plug in the value</em></u>

n + n + 2 + n + 4 + n + 6 + n + 8

507,  507 + 2 , 507 + 4 , 507 + 6 , 507 + 8

507, 509, 511, 513, 515

<u><em>Step 5:  Solve the sum of the first, third, and last integer</em></u>

507 + 511 + 515 = 1533

<u><em>Step 6: THE ANSWER</em></u>

1533

<h3><u><em>Hope this helps!!! </em></u></h3><h3><u><em>Please mark this as brainliest!!! </em></u></h3><h3><u><em>Thank You!!! </em></u></h3><h3><u><em>:)</em></u></h3>
8 0
3 years ago
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