2sin²(-2490)+tan 1410=2sin²(-330-6*360)+tan (330+3*360)=
=2sin²(-330)+tan (330)=2 sin²(30-1*360)+tan (-30)=
=2sin²(30)-tan (30)=2(1/2)²-√3/3=1/2-√3/3)=(3-2√3)/6
Answer: 2 sin²(-2490)+tan (1410)=(3-2√3)/6 (≈-0.07735...)
Answer:
7 tens
Step-by-step explanation:
the value of the first half of the statement is 83 and the value of the second one is 13 (without the tens) so that mean it has 70 left over to make it 7 tens.
Hello!
Are you looking for how many "Terms" there are?
If so, then there would be: 5 terms, which are:
a³
3a²b
-4ab
-11b
7
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