Two speakers, in phase with each other, both put out sound of frequency 260 Hz. A receiver is 2.50 m from one speaker and distan
ce x from the other, where x > 2.50 m. What is the smallest value of x such that the receiver detects maximum destructive interference
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Answer:
In this equation x = 2
Step-by-step explanation:
First, it is easiest to convert these words to numbers. That would make it
4 - <em>x </em>· 7 = -10
First subtract 4 from both sides.
4 - <em>x</em> · 7 - 4 = - 10 - 4
Then you simplify.
-<em>x ·</em> 7 = -14
Divide both sides by 7.
Then you simplify that fraction to 2 and thats how you get your answer.
Let the measure of side AB be x, then, the measue of side AE is given by
.
Now, ABCD is a square of size x, thus the area of square ABCD is given by
Also, AEFG is a square of size
, thus, the area of square AEFG is given by
<span>The sum of the areas of the two squares ABCD and AEFG is given by
Therefore, </span>the number of square units in the sum of the areas of the two squares <span>ABCD and AEFG is 81 square units.</span>
Now, ABCD is a square of size x, thus the area of square ABCD is given by
Also, AEFG is a square of size
<span>The sum of the areas of the two squares ABCD and AEFG is given by
Therefore, </span>the number of square units in the sum of the areas of the two squares <span>ABCD and AEFG is 81 square units.</span>