Melody

Well if you are in a hurry then that imples you are only really interested in the answer. 

Hence there is little reason for any of us to answer.

I am only interested in helping you to learn. I am not interested in providing you with a quick answer so that you can scrible it down, and then go off to do something  more fun.

That final answer of weighing only 50kg after 2 days is so weird.

It does seem to be right though.

Originally 

weight = 100kg   water (99%) = 99kg

after 2 days

weight = 50kg    water (98%) = 49 kg

I'll take a look :)

3. Not as important ~ I have difficulty understanding this question: 
There was a flat containing boxes of apples having a total weight of 100 kg. An analysis showed that the apples were 99% moisture, by weight. After two days in the sun, a second analysis showed that the moisture content of the apples was only 98%, by weight. What was the total weight of the apples after 2 days, in kg?

\(\text {The first analysis shows the apples are 99% water. The weight of the water is then}\\ \left(0.99\cdot 100\right) = 99 ~ kg\\ \text {Let x be the weight of the water lost after exposure to the sun. }\\ \left(0.99 \cdot 100-0.98(100-x)\right)=x\\ \)

You want this last step explained. I shall try.  

there are 100kg of apples and originally 99% of this is water so there is  0.99*100=99kg of water originally in the apples.

Over time the water dries out. All the weight that the apples lose is because of the reduced water content. 

After 2 days the water contant is only 98% of the weight. All the rest of the apple is still there.

Now in that 2 days GingerAle  has let x represents the unknown weight LOSS of the apples.

So after 2 days the apples will weigh  (100-x) kg

98% of this will be water so the new weight will be  98% of (100-x) = 0.98(100-x)

Weight after 2 days = 0.98(100-x)

So the original weight was (0.99*100)kg and the new weight is  0.98(100-x) and the weight loss is x

so

it follows that

0.99*100  -   0.98(100-x)   = x

I hope that helps.

Xsquared's  answer is the correct one.     

You do need to bew careful though.

\(1)\;\;\;-1^2=-1*1^2=-1*1=-1\\~\\ 2)\;\;\;(-1)^2=-1*-1=+1=1\\~\\ 3)\;\;\; If \;\;x=-1\;\;then\;\; x^2=(-1)^2=+1\\~\\ 4)\;\;\; x^2=1\;\;find\;\;x \qquad \rightarrow \qquad x=\pm\sqrt1=\pm1\)

That is excellent :)

\(f(x)=(2x^{3/2} -3x^{-3/2} )^2  +5\\ Find \;\;x\;\;when\;\;f(x)=5\\ (2x^{3/2} -3x^{-3/2} )^2  +5=5\\ (2x^{3/2} -3x^{-3/2} )^2  =0\\ 4x^3+9x^{-3}-12x^{0}=0\\ 4x^3+9x^{-3}-12=0\\ 4x^6+9-12x^3=0\\ 4x^6-12x^3+9=0\\ let\;\;m=x^3\\ 4m^2-12m+9=0\\ (2m-3)^2=0\\ m=\pm\left[{\frac{3}{2}}\right]^{0.5}\\ x^3=\pm\left[{\frac{3}{2}}\right]\\ \text{But x is positive so}\\ x^3=\left[{\frac{3}{2}}\right]\\ x=\left[{\frac{3}{2}}\right]^{1/3}\\ \)

1.5^(1/3) = 1.1447142425533319

x= 1.14   (to 3 significant figures)

That it the first bit, the second bit does not look too difficult, give it a try yourself.  :)

Are you sure that wasn't a +5 on the end? 

\(f(x)=(2x^{3/2} -3x^{-3/2} )^2  -5\\ Find \;\;x\;\;when\;\;f(x)=5\\ (2x^{3/2} -3x^{-3/2} )^2  -5=5\\ (2x^{3/2} -3x^{-3/2} )^2  =10\\ 4x^3+9x^{-3}-12x^{0}=100\\ 4x^3+9x^{-3}-12x=100\\ 4x^6+9-12x^4=100x^3\\ 4x^6-12x^4-100x^3+9=0\\ \)

Thanks guest :)

Dont be so lazy.

Type your question in, it is only words. 

Here are the 4 extra acute angles