I think it’s four. If not then I’m sorry

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

.
Answer:
68
Step-by-step explanation:
1) Plug In (when you plug in you replace the defined variable with the number given. usually you have to find the numbers to plug in but since they're already given to you you just plug them in.) 4(3) + 8(7)=T T= total cost
2) Multiply 4*3=12 8*7=56
3) Add 12+56=68
4) <u>T=68</u>
<2= 30 degrees since its vertical to the 4th angle.
<1=150 degrees
<3=150 degrees
since angle 3 and the 4th angle are supplementary, which means both angles equal 180 degrees, angle 3 equals 150 degrees.
since angle 3 is vertical to angle 1 it's also 150 degrees since they're congruent.
The answer is c because 2+7=9 and 10-2=8 and 9 is greater than 8 so 2+7>10-2