Can someone help me out with number 2?

Mathematics

1 answer:

Norma-Jean [14]2 years ago

5 0

$\sin255^o=\sin(300^o-45^o)=\sin300^o\cos45^o-\cos300^o\sin45^o= \\\\=- \frac{ \sqrt{3} }{2}\cdot \frac{ \sqrt{2} }{2}- \frac{1}{2} \cdot \frac{ \sqrt{2} }{2}=- \frac{ \sqrt{6} }{4} - \frac{ \sqrt{2} }{4}= -\frac{ \sqrt{6}+\sqrt{2} }{4}$

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A 12,000-lb killer whale might eat as much as 14,000 lb of fish per month (assume a 30-day month). If the average weight of a ce

brilliants [131]

Answer:

2196 fish

Step-by-step explanation:

Weight of fish that the killer whale might eat in a month (30 days) = 14,000 lb

Weight of the fish that killer whale might eat in a day = $\frac{14000}{30} = \frac{1400}{3}$ lb.

Average weight of the fish that killer whale eats = 3.4 oz

Since,

1 pound(lb) = 16 ounces(oz)

So,

Weight of the fish that killer whale might eat in a day in ounces(oz) = $\frac{1400}{3} \times 16=\frac{22400}{3}$ oz

Since, a certain fish weighs 3.4 oz, in order to find the number of fish that killer whale must eat, we must divide the total weight in ounces by 3.4 ounces.

So, the number of fish that killer whale would eat = $\frac{22400}{3} \div 3.4 = 2196$ fish