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Gelneren [198K]
3 years ago
6

Homer Simpson weighs 400 pounds. He is on a diet to lose 5 pounds each week Barney weighs 322 pounds. he is on a diet to lose 3

pounds each week. How long will it take for them to weigh the same? write an equation and solve.
Mathematics
1 answer:
Triss [41]3 years ago
3 0
400 + (n-1) (-5) = 322 + (n-1) (-3)
-5n + 5 = 322 
-3n + 3    | 405 - 5n
               =325 - 3n
80=2n
n=40            n - 1 = 39    

 Answer = 39 , 39 weeks until they weigh the same
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Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) \cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}

( 2 ) \sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}

These two identities makes sin(π / 10) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and cos(π / 10) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}.

Therefore cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}. Substitute,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right]

And now simplify this expression to receive our answer,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right] = -\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i,

-\frac{3\sqrt{5+\sqrt{5}}}{4} = -2.01749\dots and \:\frac{3\sqrt{3-\sqrt{5}}}{4} = 0.65552\dots

= -2.01749+0.65552i

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}, cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}

We know that 6\sqrt{5+\sqrt{5}} = 16.13996\dots and -\:6\sqrt{3-\sqrt{5}} = -5.24419\dots . Therefore,

Solution : 16.13996 - 5.24419i

Which rounds to about option b.

7 0
3 years ago
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