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Olegator [25]
2 years ago
12

Last Saturday, Camila practiced her dance routines for her upcoming competition. She spent some time practicing her tap routine,

and then she spent 17 minutes practicing her hip-hop routine. In all, she practiced for 45 minutes. Let t be the number of minutes that Camila practiced her tap routine. Write and solve an equation to find t.
Mathematics
1 answer:
Natalka [10]2 years ago
4 0

Answer:

t = 28 minutes

Step-by-step explanation:

Last Saturday, Camila practiced her dance routines for her upcoming competition. She spent some time practicing her tap routine, and then she spent 17 minutes practicing her hip-hop routine. In all, she practiced for 45 minutes. Let t be the number of minutes that Camila practiced her tap routine. Write and solve an equation to find t.

45 minutes = Number of minutes for hip-hop routine + Number of minutes for tap dance

The equation is given as

45 = 17 + t

Solving for t

t = 45 - 17

t = 28 minutes

Shes spent 28 minutes tap dancing

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Explanation:

The key to area in polar coordinates is the formula for the area of a sector:

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Then a differential of area* can be written as ...

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Filling in the given function for r, we have ...

  da = (1/2)(4cos(3θ))²·dθ = 8cos(3θ)²·dθ

The integral will have limits corresponding to the range of values of θ for one loop of the graph: -π/6 to π/6. So, the area is ...

  \displaystyle A=\int{da}=\int\limits_{-\pi /6}^{\pi /6}{8\cos{(3\theta)}^2}\, d\theta\\\\=8\int\limits_{-\pi /6}^{\pi /6}{\cos{(3\theta)}^2}

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* As with other approaches to finding area (horizontal or vertical slice, for example), we assume that the differential element dθ is sufficiently small that we need not concern ourselves with the fact that r is a function of θ.

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The ordered pairs (3 , 6) , (5 , 10) show a proportional relationship ⇒ last answer

Step-by-step explanation:

* Lets explain how to sole the problem

- Proportional relationship describes a simple relation between

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- In direct proportion if one variable increases, then the other variable

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- In inverse proportion if one variable increases, then the other variable

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* Lets solve the problem

# Last table

∵ x = 3 and y = 6

∴ x/y = 3/6 = 1/2

∵ x = 5 and y = 10

∴ x/y = 5/10 = 1/2

∵ 1/2 is constant

∵ x/y = constant

∴ x and y are proportion

* The ordered pairs (3 , 6) , (5 , 10) show a proportional relationship

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An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem
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DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

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