Rule 1: The product<span> of a positive </span>integer<span> and a negative </span>integer<span> is a negative </span>integer<span>. Rule 2: The </span>product of two<span> negative </span>integers<span> or </span>two<span> positive </span>integers<span> is a positive </span><span>integer</span>
The answer would be the third one
by the double angle identity for sine. Move everything to one side and factor out the cosine term.
Now the zero product property tells us that there are two cases where this is true,
In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of
, so
where
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](https://tex.z-dn.net/?f=n%5B%2Ftex%20%5Dis%20any%20integer.%5C%5CMeanwhile%2C%5C%5C%5Btex%5D10%5Csin%20x-3%3D0%5Cimplies%5Csin%20x%3D%5Cdfrac3%7B10%7D)
which occurs twice in the interval
for
and
. More generally, if you think of
as a point on the unit circle, this occurs whenever
also completes a full revolution about the origin. This means for any integer
, the general solution in this case would be
and
.
150
13+46= 59
59+87= 146
Round 46 to the nearest ten, which in this case would be 50 because the last number is larger than a 5.
Then add the 1 back in to make it 150.
Answer:
700
Step-by-step explanation:
