sin(3)+cos(3)sin(3)tan(3)3=0=−cos(3)=−1=−4+,∈ℤ=−12+3
sin(3x)+cos(3x) =0 sin(3x) =−cos(3x) tan(3x) =−1 3x =−
π
4
+kπ,k∈
Z
x =−
π
12
+
k
π
3
Since −<<
−
π
<
x
<
π
, −2≤≤3
−
2
≤
k
≤
3
. Thus, the solution set is
{−34,−512,−12,4,712,1112}
3.
You could probably look up a fancy explanation of why this is, by I just went the easy route.
1 is definitely possible.
2 is too!
(Sorry if the images are hard to see)
So the answer has got to be 3!
Hope this helps!
The answer would be 280 as a tens bar is worth ten. 28x10=280
A + B + C = 476
A = 3B + 6
C = B + 45
now its just a matter of subbing..
A + B + C = 476
(3B + 6) + B + (B + 45) = 476...combine like terms
5B + 51 = 476
5B = 476 - 51
5B = 425
B = 425/5
B = 85 <== team B scored 85
A = 3B + 6
A = 3(85) + 6
A = 255 + 6
A = 261 <=== team A scored 261
C = B + 45
C = 85 + 45
C = 130 <=== team C scored 130
950ml × 1L1000ml = 0.95L is the correct answer.