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pishuonlain [190]
2 years ago
8

You want to make a necklace with 5 beads on it; you have a bag that contains 5 differently colored beads. You draw a bead from t

he bag and string it on the necklace. Then you draw another bead and add it to the necklace. This process is repeated until all of the beads are used. How many different color combinations are possible for the necklace?
Mathematics
1 answer:
Vladimir [108]2 years ago
8 0
There are 120 different combinations.

There are 5 to use for the first bead; then after it is used, 4 for the second; 3 for the third; 2 for the fourth; and 1 for the fifth and last bead:

5*4*3*2*1 = 120.
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arsen [322]

Answer:

the answer is 32 :)

hope this helps

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Simon put $15000 in an account with a simple interest rate of 4.5%. How much interest will be earned after 8 years?
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~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$15000\\ r=rate\to 4.5\%\to \frac{4.5}{100}\dotfill &0.045\\ t=years\dotfill &8 \end{cases} \\\\\\ I = (15000)(0.045)(8)\implies I=5400

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2 years ago
8/9=x+1/3(7) how to solve
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Do 1/3 x 7 (due to the parentheses which mean multiplication). Once you get that answer you should have 8/9 = x + (answer). Now subtract that answer from 8/9 and that's your x!
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3 years ago
There are 6 red marbles, 8 blue marbles, and 11 green marbles in a bag. What is the probability that you will randomly draw eith
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<span>6/25 + 8/25 =0.56
</span><span>0.56 or as a percent 56%SO the answer is B 56%
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Can someone help me out !<br><br>got stuck in this question for an hour.<br><br><br>​
Reil [10]

Answer:

See below

Step-by-step explanation:

Considering $\vec{u}, \vec{v}, \vec{w} \in V^3 \lambda \in \mathbb{R}$, then

\Vert \vec{u} \cdot \vec{v}\Vert \leq  \Vert\vec{u}\Vert  \Vert\vec{v}\Vert$ we have $(\vec{u} \cdot \vec{v})^2 \leq (\vec{u} \cdot \vec{u})(\vec{v} \cdot \vec{v}) \quad$

This is the Cauchy–Schwarz  Inequality, therefore

$\left(\sum_{i=1}^{n} u_i v_i \right)^2 \leq \left(\sum_{i=1}^{n} u_i \right)^2 \left(\sum_{i=1}^{n} v_i \right)^2  $

We have the equation

\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b}  = \dfrac{1}{a+b}, a,b\in\mathbb{N}

We can use the Cauchy–Schwarz  Inequality because a and b are greater than 0. In fact, a>0 \wedge b>0 \implies ab>0. Using the Cauchy–Schwarz  Inequality, we have

\dfrac{\sin ^4 x }{a} + \dfrac{\cos^4 x }{b}   =\dfrac{(\sin^2 x)^2}{a}+\dfrac{(\cos^2 x)}{b}\geq \dfrac{(\sin^2 x+\cos^2 x)^2}{a+b} = \dfrac{1}{a+b}

and the equation holds for

\dfrac{\sin^2{x}}{a}=\dfrac{\cos^2{x}}{b}=\dfrac{1}{a+b}

\implies\quad \sin^2 x = \dfrac{a}{a+b} \text{ and }\cos^2 x = \dfrac{b}{a+b}

Therefore, once we can write

\sin^2 x = \dfrac{a}{a+b} \implies \sin^{4n}x = \dfrac{a^{2n}}{(a+b)^{2n}} \implies\dfrac{\sin^{4n}x }{a^{2n-1}} = \dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}

It is the same thing for cosine, thus

\cos^2 x = \dfrac{b}{a+b} \implies \dfrac{\cos^{4n}x }{b^{2n-1}} = \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}}

Once

\dfrac{a^{2n}}{(a+b)^{2n}\cdot a^{2n-1}}+ \dfrac{b^{2n}}{(a+b)^{2n}\cdot b^{2n-1}} =\dfrac{a^{2n}}{(a+b)^{2n} \cdot \dfrac{a^{2n}}{a} } + \dfrac{b^{2n}}{(a+b)^{2n}\cdot \dfrac{b^{2n}}{b} }

=\dfrac{1}{(a+b)^{2n} \cdot \dfrac{1}{a} } + \dfrac{1}{(a+b)^{2n}\cdot \dfrac{1}{b} } = \dfrac{a}{(a+b)^{2n}  } + \dfrac{b}{(a+b)^{2n} } = \dfrac{a+b}{(a+b)^{2n} }

dividing both numerator and denominator by (a+b), we get

\dfrac{a+b}{(a+b)^{2n} } =  \dfrac{1}{(a+b)^{2n-1} }

Therefore, it is proved that

\dfrac{\sin ^{4n} x }{a^{2n-1}} + \dfrac{\cos^{4n} x }{b^{2n-1}}  = \dfrac{1}{(a+b)^{2n-1}}, a,b\in\mathbb{N}

4 0
3 years ago
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