Find the opposite of adding 15. So it would be subtracting 15. So subtract 15 on both sides. So it would have 2/3x on the left and then 2 on the other side. (2/3x=2) Now find the opposite of multiplying 2/3. It is dividing 23/3. So divide 23/3 on both sides. So multiply 2/3 times 1/2. It would equal 3. x = 3
Hope this helps. :-)
Answer:
98.5 cm2
Step-by-step explanation:
Find <span>tan<span>(<span><span>5π</span>12</span>)</span></span> and sin ((5pi)/12)
Answer: <span>±<span>(2±<span>√3</span>)</span>and±<span><span>√<span>2+<span>√3</span></span></span>2</span></span>
Explanation:
Call tan ((5pi/12) = t.
Use trig identity: <span><span>tan2</span>a=<span><span>2<span>tana</span></span><span>1−<span><span>tan2</span>a</span></span></span></span>
<span><span>tan<span>(<span><span>10π</span>12</span>)</span></span>=<span>tan<span>(<span><span>5π</span>6</span>)</span></span>=−<span>1<span>√3</span></span>=<span><span>2t</span><span>1−<span>t2</span></span></span></span>
<span><span>t2</span>−2<span>√3</span>t−1=0</span>
<span>D=<span>d2</span>=<span>b2</span>−4ac=12+4=16</span>--> <span>d=±4</span>
<span>t=<span>tan<span>(<span><span>5π</span>12</span>)</span></span>=<span><span>2<span>√3</span></span>2</span>±<span>42</span>=2±<span>√3</span></span>
Call <span><span>sin<span>(<span><span>5π</span>12</span>)</span></span>=<span>siny</span></span>
Use trig identity: <span><span>cos2</span>a=1−2<span><span>sin2</span>a</span></span>
<span><span>cos<span>(<span><span>10π</span>12</span>)</span></span>=<span>cos<span>(<span><span>5π</span>6</span>)</span></span>=<span><span>−<span>√3</span></span>2</span>=1−2<span><span>sin2</span>y</span></span>
<span><span><span>sin2</span>y</span>=<span><span>2+<span>√3</span></span>4</span></span>
<span><span>siny</span>=<span>sin<span>(<span><span>5π</span>12</span>)</span></span>=±<span><span><span>√<span>2+<span>√3</span></span></span>2</span></span></span>
44.2%= 287
Let the total attempt = x
(287/x) ×100 = 44.2
287/x = 44.2/100
287/x = 0.442
x = 287/0.442
<h3>x= <u>649 </u></h3>
Hope this will help...
