To solve this problem you must apply the proccedure shown below:
1. You have to r<span>ewrite x=12 in polar form. Then, you have:
12=rCos</span><span>θ
2. Then, you must solve for r, as following:
r=12/Cos</span><span>θ
</span> 3. You have that 1/Cosθ=Sec<span>θ, therefore:
</span> r=12(1/Cos<span>θ)
</span> r=12Sec<span>θ
</span> Therefore, as you can see, the answer is: r=12Secθ<span>
</span>
1) The outcomes for rolling two dice, the sample space, is as follows:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
There are 36 outcomes in the sample space.
2) The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5). There are 18 outcomes in this event.
3) The probability of rolling an odd sum is 18/36 = 1/2 = 0.5
If every 5 mins, A makes 1 yo-yo every 10 mins, B makes 1 yo-yo then every 10 mins, both machines produce 3 yo-yos every 10 mins (2 from machine A and 1 from machine B) Therefore, for 20 yo-yos, both machines would take 70 minutes( 1 hour and 10 mins). After 70 minutes, 21 yo-yos would be produced.
