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Vladimir [108]
3 years ago
12

A large box of jelly beans weighs 96.58 lb the jelly beans are evenly divided into 22 bags what is the weight in pounds of the j

elly beans in each bag
Mathematics
1 answer:
Vadim26 [7]3 years ago
4 0

Answer:

The weight in each bag is 4.39lb.

Step-by-step explanation:

To solve this, you simply need to divide 96.58 by 22. The reason you do this is so that way you can divide all of the weight into 22 bags.

96.58 ÷ 22 = 4.39

This means that the weight in each bag is 4.39lb.

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Simplify using the distributive property. 5 Marks<br> 1/5(4m+20)<br><br> 9(5/9 y-1/3)
Zarrin [17]

Answer+Step-by-step explanation:

\frac{1}{5} \left( 4m+20\right)  =\frac{4}{5}m + \frac{20}{5} = \frac{4}{5}m + 4

9\left( \frac{5}{4} y-\frac{1}{3} \right)  =\frac{9\times5}{9} y-\frac{9}{3} = 5y - 3

4 0
2 years ago
Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b
trapecia [35]

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.

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now, if the denominator turns to 0, the fraction becomes undefined, and you get a "vertical asymptote" when that happens, so let's check when is that

\bf sin\left(x-\frac{2\pi }{3}  \right)=0\implies sin^{-1}\left[ sin\left(x-\frac{2\pi }{3}  \right) \right]=sin^{-1}(0)&#10;\\\\\\&#10;x-\frac{2\pi }{3}=&#10;\begin{cases}&#10;0\\&#10;\pi &#10;\end{cases}\implies \measuredangle x=&#10;\begin{cases}&#10;\frac{2\pi }{3}\\&#10;\frac{5\pi }{3}&#10;\end{cases}

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