Well, you can do that by many ways like:
If you have value of B & P or B & H, you can find the other one by the formula,
H² = P²+B²
Then, you can easily calculate the value of sine by putting P/H

Hope this helps!

4 0

2 years ago

Read 2 more answers

Six rectangles each with a common base width of 22 have lengths of 1,4,9,16,251,4,9,16,25, and 3636. What is the sum of the area

Angelina_Jolie [31]

22(1 + 4 + 9 + 16 +251 + 4 + 9 + 16 + 25 + 3636)
= 87362 units^2

8 0

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 y 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 |x | ≤ 605.

⇒ -605 ≤ x ≤ 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.

7 0

2 years ago

There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit

Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a $\frac{3}{11}$ chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is $\frac{8}{11}$ , and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

$\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}$

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is $\frac{2}{11}$ , and the probability of picking the second red ball is $\frac{1}{10}$ and the probability of picking the green ball is $\frac{1}{9}$ . We want to multiply thm again, as per the multiplication rule like the last problem.

$\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}$

5 0

2 years ago