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2 years ago

If the pattern continues how many cups of milk does George need to make 6 batches of muffins

Mathematics

1 answer:

alexdok [17]2 years ago

6 0

Answer:

see explanation

Step-by-step explanation:

If we consider the differences between consecutive terms, that is

$\frac{3}{4}$ - $\frac{3}{8}$ = $\frac{3}{8}$

1 $\frac{1}{8}$ - $\frac{3}{4}$ = $\frac{3}{8}$

1 $\frac{1}{2}$ - 1 $\frac{1}{8}$ = $\frac{3}{8}$

To obtain the next term in the pattern add $\frac{3}{8}$ to the previous term.

batches = 5 ⇒ milk = 1 $\frac{1}{2}$ + $\frac{3}{8}$ = 1 $\frac{7}{8}$

batches = 6 ⇒ milk = 1 $\frac{7}{8}$ + $\frac{3}{8}$ = 2 $\frac{1}{4}$

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The probability of spinning vanilla ice cream with chocolate a rainbow sprinkles is 1/12

<h3>The missing part of the question</h3>

Flavor of ice cream: Vanilla, Strawberry, Chocolate

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<h3>The number of different combinations</h3>

Using the dataset above, we have:

Ice cream = 3

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The number of different combinations is calculated as:

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Combination = 12

Hence, the number of different combinations is 12

<h3>The probability of spinning vanilla ice cream with chocolate a rainbow sprinkles</h3>

The vanilla ice cream with chocolate a rainbow sprinkles is just one of the 12 combinations.

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1 year ago

Find $\int \dfrac{x}{\sqrt{1-x^4}}$ Please, help

ki77a [65]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-x^4}}\,dx}\\\\\\ \mathsf{=\displaystyle\int\! \frac{1}{2}\cdot 2\cdot \frac{1}{\sqrt{1-(x^2)^2}}\,dx}\\\\\\ \mathsf{=\displaystyle \frac{1}{2}\int\! \frac{1}{\sqrt{1-(x^2)^2}}\cdot 2x\,dx\qquad\quad(i)}$

Make a trigonometric substitution:

$\begin{array}{lcl}
\mathsf{x^2=sin\,t}&\quad\Rightarrow\quad&\mathsf{2x\,dx=cos\,t\,dt}\\\\
&&\mathsf{t=arcsin(x^2)\,,\qquad 0\ \textless \ x\ \textless \ \frac{\pi}{2}}\end{array}$

so the integral (i) becomes

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{1-sin^2\,t}}\cdot cos\,t\,dt\qquad\quad (but~1-sin^2\,t=cos^2\,t)}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{\sqrt{cos^2\,t}}\cdot cos\,t\,dt}$

$\mathsf{=\displaystyle\frac{1}{2}\int\!\frac{1}{cos\,t}\cdot cos\,t\,dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\int\!\f dt}\\\\\\
\mathsf{=\displaystyle\frac{1}{2}\,t+C}$

Now, substitute back for t = arcsin(x²), and you finally get the result:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=\frac{1}{2}\,arcsin(x^2)+C}$ ✔

________

You could also make

x² = cos t

and you would get this expression for the integral:

$\mathsf{\displaystyle\int\! \frac{x}{\sqrt{1-(x^2)^2}}\,dx=-\,\frac{1}{2}\,arccos(x^2)+C_2}$ ✔

which is fine, because those two functions have the same derivative, as the difference between them is a constant:

$\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\
=\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\
=\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}$

$\mathsf{=\dfrac{\pi}{4}}$ ✔

and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.

I hope this helps. =)

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