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irina [24]
2 years ago
5

At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the firs

t two races. If the first race runs 9 horses and the second runs 8, how many different tickets are possible
Mathematics
1 answer:
rewona [7]2 years ago
8 0

Answer:

4032 different tickets are possible.

Step-by-step explanation:

Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.

To find : How many different tickets are possible ?

Solution :

In the first race there are 9 ways to pick the winner for first and second place.

Number of ways for first place - ^9C_1=9

Number of ways for second place - ^8C_1=8

In the second race there are 8 ways to pick the winner for first and second place.

Number of ways for first place - ^8C_1=8

Number of ways for second place - ^7C_1=7

Total number of different tickets are possible is

n=9\times 8\times 8\times 7

n=4032

Therefore, 4032 different tickets are possible.

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Find side x , give your answer 1 decimal place <br>​
Vladimir [108]

<u>Given</u>:

Let A denote the angle of the given triangle which measures ∠A = 81°

Let B denote the angle which measures ∠B = 40°

The length of the side a = x.

The length of the side b = 7 cm.

We need to determine the value of x.

<u>Value of x:</u>

The value of x can be determined using the law of sine formula.

Applying the law of sine, we have;

\frac{sin \ A}{a}=\frac{sin \ B}{b}

Substituting the values, we have;

\frac{sin \ 81^{\circ}}{x}=\frac{sin \ 40^{\circ}}{7}

Simplifying, we get;

\frac{0.988}{x}=\frac{0.643}{7}

Cross multiplying, we have;

0.988 \times 7=0.643 x

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Thus, the length of the side x is 10.8 cm.

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2 years ago
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2 years ago
The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and
brilliants [131]

Answer:

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Step-by-step explanation:

The equation of the parabola is:

y=0.00035x^{2}

Compute the first order derivative of <em>y</em> as follows:

 y=0.00035x^{2}

\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]

    =2\cdot 0.00035x\\\\=0.0007x

Now, it is provided that |<em>x </em>| ≤ 605.

⇒ -605 ≤ <em>x</em> ≤ 605

Compute the arc length as follows:

\text{Arc Length}=\int\limits^{x}_{-x} {1+(\frac{\text{dy}}{\text{dx}})^{2}} \, dx

                  =\int\limits^{605}_{-605} {\sqrt{1+(0.0007x)^{2}}} \, dx \\\\={\displaystyle\int\limits^{605}_{-605}}\sqrt{\dfrac{49x^2}{100000000}+1}\,\mathrm{d}x\\\\={\dfrac{1}{10000}}}{\displaystyle\int\limits^{605}_{-605}}\sqrt{49x^2+100000000}\,\mathrm{d}x\\\\

Now, let

x=\dfrac{10000\tan\left(u\right)}{7}\\\\\Rightarrow u=\arctan\left(\dfrac{7x}{10000}\right)\\\\\Rightarrow \mathrm{d}x=\dfrac{10000\sec^2\left(u\right)}{7}\,\mathrm{d}u

\int dx={\displaystyle\int\limits}\dfrac{10000\sec^2\left(u\right)\sqrt{100000000\tan^2\left(u\right)+100000000}}{7}\,\mathrm{d}u

                  ={\dfrac{100000000}{7}}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=\dfrac{50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{7}+\dfrac{50000000\sec\left(u\right)\tan\left(u\right)}{7}\\\\=\dfrac{50000000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+5000x\sqrt{\dfrac{49x^2}{100000000}+1}

Plug in the solved integrals in Arc Length and solve as follows:

\text{Arc Length}=\dfrac{5000\ln\left(\sqrt{\frac{49x^2}{100000000}+1}+\frac{7x}{10000}\right)}{7}+\dfrac{x\sqrt{\frac{49x^2}{100000000}+1}}{2}|_{limits^{605}_{-605}}\\\\

                  =1245.253707795227\\\\\approx 1245.25

Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

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