Determine the amplitude or period as requested. Amplitude of y = 1/2 * cos 5x
1 answer:
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Answer:
<em><u>1=</u></em><em><u> </u></em> 237
<em><u>2=</u></em> 200
<em><u>3</u></em><em>= </em>49
<em><u>4=</u></em> 50
<em>I'm </em><em>99% </em><em>sure these are right.</em>
<em> --------------------------------------------------------</em>
<em> </em><u><em>please let me know if </em></u><u>any are wrong.</u>
<h2> Have a good day<3</h2><h2></h2><h2> </h2>
Okay so the y is in absolute value, which means the number could be a negative and the absolute value would make is a positive.
since the sign is greater than, with no equal to underneath it, than the answer cant be 3 or -3.
the number has to total to be larger than 3. again, since the absolute value turns the negatives into positives, then -4 is an acceptable answer. so the absolute value line should look like the above (please excuse the terrible look). I simplified it to the two numbers that start the equation. the two circle are open circles because the sign does not include an equal sign under it. you draw the arrows on either side of the two numbers because any answer beyond those arrows fits in with the equation
| 4 |
| 184728 |
| -4.5829492994929492949572728 |
did that answer the problem?
since the sign is greater than, with no equal to underneath it, than the answer cant be 3 or -3.
the number has to total to be larger than 3. again, since the absolute value turns the negatives into positives, then -4 is an acceptable answer. so the absolute value line should look like the above (please excuse the terrible look). I simplified it to the two numbers that start the equation. the two circle are open circles because the sign does not include an equal sign under it. you draw the arrows on either side of the two numbers because any answer beyond those arrows fits in with the equation
| 4 |
| 184728 |
| -4.5829492994929492949572728 |
did that answer the problem?
Let s be the side length of the square. The dimensions of the rectangle are three times the side of the square (i.e. 3s), and two less than the side of the square (i.e. s-2).
So, the area of this rectangle is
The area of the square is , and we know that the two areas are the same, so we have
The solution s=0 would lead to the extreme case where the rectangle and the square are actually a point, so we accept the solution s=3.