Steps:
Find the 2D area of the two triangles:
- multiply 6 x 8 and divide it by 2 (this is the bottom triangle.)
** don't use 10 because it is the hypotenuse.
- 6 x 8 = 48
- 48 / 2 = 24 (you will need to use this twice since there are 2 triangles)
Now find the 2D area of the three rectangles
- first you have to convert 4 yd to feet.
**1 yd = 3 ft
~4 yd = 12 ft
-now multiply 12 x 10
- 12 x 10 = 120ft (use this 3 times since there are 3 rectangles)
Now add all those together:
120 + 120 + 120 + 24 + 24 = 408
90cm:1.5m=90cm:150cm=9:15=3:5=3/5
-1/4 cos 2x + C
<span>or </span>1/2 sin^2 x + C
or -1/2 cos^2 x + C
well, <span><span>sinx</span><span>cosx</span>=<span><span><span>sin2</span>x</span>2</span></span> so you are looking at<span><span>12</span>∫ <span>sin2</span>x <span>dx</span>=<span>(<span>12</span>)</span><span>[<span>(<span>12</span>)</span><span>(−<span>cos2</span>x)</span>+C]</span>=−<span>14</span><span>cos2</span>x+C'</span>
or maybe easier you can notice the pattern that <span><span>(<span><span>sinn</span>x</span>)</span>'=n<span><span>sin<span>n−1</span></span>x</span><span>cosx</span></span> and pattern match. here <span>n−1=1</span> so n = 2 so we trial <span><span>(<span><span>sin2</span>x</span>)</span>'</span> which gives us <span>2<span>sinx</span><span>cosx</span></span> so we now that the anti deriv is <span><span>12</span><span><span>sin2</span>x</span>+C</span>
the other pattern also works ie<span><span>(<span><span>cosn</span>x</span>)</span>'=n<span><span>cos<span>n−1</span></span>x</span><span>(−<span>sinx</span>)</span>=−n<span><span>cos<span>n−1</span></span>x</span><span>sinx</span></span>
so trial solution <span><span>(−<span><span>cos2</span>x</span>)</span>'=−2<span>cosx</span><span>(−<span>sinx</span>)</span>=2<span>cosx</span><span>sinx</span></span> so the anti deriv is <span>−<span>12</span><span><span>cos2</span>x</span>+<span>C</span></span>
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Hello!
It cost an adult $15 to attend a show and $10 for a child.
Hope this helps!
Answer: (-3,-4x)
Solve for the first variable in one of the equations, then substitute the results into the other equation.