Randy took out a subsidized student loan of $7,000 at a 3.6% APR, compounded monthly, to pay for his last semester of college. I
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<h3>Answer:</h3>
- using y = x, the error is about 0.1812
- using y = (x -π/4 +1)/√2, the error is about 0.02620
The actual value of sin(π/3) is (√3)/2 ≈ 0.86602540.
If the sine function is approximated by y=x (no error at x = 0), then the error at x=π/3 is ...
... x -sin(x) @ x=π/3
... π/3 -(√3)/2 ≈ 0.18117215 ≈ 0.1812
You know right away this is a bad approximation, because the approximate value is π/3 ≈ 1.04719755, a value greater than 1. The range of the sine function is [-1, 1] so there will be no values greater than 1.
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If the sine function is approximated by y=(x+1-π/4)/√2 (no error at x=π/4), then the error at x=π/3 is ...
... (x+1-π/4)/√2 -sin(x) @ x=π/3
... (π/12 +1)/√2 -(√3)/2 ≈ 0.026201500 ≈ 0.02620
<u>the </u><u>given </u><u>figure </u><u>is </u><u>a </u><u>composition</u><u> </u><u>of </u><u>a </u><u>rectangle</u><u> </u><u>as </u><u>well </u><u>as </u><u>a </u><u>right </u><u>angled </u><u>triangle </u><u>!</u>
<u>we've</u><u> </u><u>been </u><u>given </u><u>the </u><u>two </u><u>sides </u><u>of </u><u>the </u><u>rectangle </u><u>and </u><u>we're</u><u> </u><u>required</u><u> </u><u>to </u><u>find </u><u>out </u><u>the </u><u>height </u><u>of </u><u>the </u><u>triangle </u><u>,</u><u> </u><u>so </u><u>as </u><u>to </u><u>find </u><u>it's</u><u> </u><u>area </u><u>~</u>
<u>we </u><u>know </u><u>the </u><u>the </u><u>opposite</u><u> </u><u>sides </u><u>of </u><u>a </u><u>rectangle </u><u>are </u><u>equal</u><u> </u><u>,</u><u> </u><u>therefore </u><u>we </u><u>can </u><u>break </u><u>the </u><u>longest </u><u>side </u><u>(</u><u> </u><u>length </u><u>=</u><u> </u><u>9</u><u>.</u><u>5</u><u> </u><u>cm </u><u>)</u><u> </u><u>into </u><u>two </u><u>parts </u><u>!</u><u> </u><u>the </u><u>first </u><u>part </u><u>of </u><u>length </u><u>=</u><u> </u><u>7</u><u> </u><u>cm </u><u>which </u><u>is </u><u>the </u><u>length </u><u>of </u><u>the </u><u>rectangle </u><u>and </u><u>the </u><u>rest </u><u>2</u><u>.</u><u>5</u><u> </u><u>cm </u><u>(</u><u> </u><u>9</u><u>.</u><u>5</u><u> </u><u>-</u><u> </u><u>7</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>)</u><u> </u><u>will </u><u>become </u><u>the </u><u>height </u><u>of </u><u>the </u><u>triangle </u><u>!</u>
<h3><u>For </u><u>perimeter</u><u> </u><u>of </u><u>the </u><u>figure </u><u>-</u></h3>now ,
<u>perimeter</u><u> </u><u>of </u><u>rectangle </u><u>=</u><u> </u><u>2</u><u> </u><u>(</u><u> </u><u>l </u><u>+</u><u> </u><u>b </u><u>)</u>
where ,
<u>l </u><u>=</u><u> </u><u>length </u>
<u>b </u><u>=</u><u> </u><u>breadth </u>
and ,
<u>Perimeter</u><u> </u><u>of </u><u>figure </u><u>in </u><u>total </u><u>=</u><u> </u><u>2</u><u>6</u><u> </u><u>cm </u><u>+</u><u> </u><u>1</u><u>5</u><u> </u><u>cm</u>
thus ,
now ,
<u>area </u><u>of </u><u>rectangle</u><u> </u><u>=</u><u> </u><u>l </u><u>×</u><u> </u><u>b</u>
where ,
<u>l </u><u>=</u><u> </u><u>length </u>
<u>b </u><u>=</u><u> </u><u>breadth</u>
and ,
<u>Area </u><u>of </u><u>figure</u><u> </u><u>in </u><u>total </u><u>=</u><u> </u><u>4</u><u>2</u><u> </u><u>cm²</u><u> </u><u>+</u><u> </u><u>7</u><u>.</u><u>5</u><u> </u><u>cm²</u>
thus ,
hope helpful :)