The new period is <span>
2/3 π</span>
.
The period of the two elementary trig functions, <span>sin<span>(x)</span></span><span> and </span><span>cos<span>(x)</span></span><span> is </span><span>2π</span><span>.
</span>
If we multiply the input variable by a constant has the effect of stretching or contracting the period. If the constant, c>1 then the period is stretched, if c<1 then the period is contracted.
We can see what change has been made to the period, T, by solving the equation:
<span>cT=2π</span>
What we are doing here is checking what new number, T, will effectively input the old period, 2π, to the function in light of the constant. So for our givens:
<span>3T=2π</span>
<span>T=2/3 π</span>
Other method to solve this;
<span><span>sin3</span>x=<span>sin<span>(3x+2π)</span></span>=<span>sin<span>[3<span>(x+<span><span>2π/</span>3</span>)</span>]</span></span>=<span>sin3</span>x</span>
This means "after the arc rotating three time of <span>(x+<span>(2<span>π/3</span>)</span>)</span>, sin 3x comes back to its initial value"
So, the period of sin 3x is <span><span>2π/</span>3 or 2/3 </span>π.
Its 0.9 or 0.90 thats only i know
Add all the numbers: 33,37,&41 then divide 111 by 3 and u will get 37.
Answer:
C
Step-by-step explanation:
If you have a Ti-84 series calculator, press "stat" then "Edit..." and then fill in the data table values for x and y in two lists. Then press "2nd" and "mode" to quit. Now press "stat" again and right arrow over to "calc" and press down until you find "ExpReg" and set the "Xlist" and "Ylist" that you used and you will get C as the answer. Another way to do this is to manually substitute values into all 4 equations, which is boring.