amygdaleon305

Let's assume that Chloe and Sam had \(x\) apples at first.

So then, 

\(x+3={60 \over 100}(x+23)\)

\(100x+300=60x+1380\)

\(40x=1080\)

⇒\(x=27\)

∴ Sam had 27 apples at first. 

Let's assume Ahmed had \(x\) at a first. 

∴ Brad had = \({14x \over 5}\)

Now, according to question

\({14x \over 5}-20=x+16\)

\({14x \over 5}-x=16+20\)

\(9x=180\)

\(x=20\)

∴ Ahmed had 20 at first. 

Its given that 

\(f(x)= ax^3-6x^2+bx-5\)

When f(x) is divided by x-1 we get remainder 

\(r=b-a+1\)

⇒\(b-a+1=-5\)

⇒\(a-b=6\)                                           ...(1)                           

Now, when f(x) is divided by x+2 we get remainder 

\(r' = -8a-2b-29\)

⇒\(-8a-2b-29=53\)

⇒\(-8a-2b=82\)

⇒\(-4a-b=41\)                                   ...(2)

Subtracting eq (1) from (2)

\(-4a-b-a+b=41-6\)

\(-5a=35\)

∴ \(a=-7\)

From eq (1)

\(b=-13\)

Hence the ordered pair (a,b) is (-7,-13). 

~Thank You :)

Let's assume the integers are a-2, a-1, a, a+1, a+2.

Now, according to question 

⇒\(a(a+2) =483\)

⇒\(a^2+2a-483 = 0\)

⇒\((a+23)(a-21)=0\)

∴ a = -23 or 21

Obviously, since a is negative,

⇒ a = -23

∴ 5 consecutive negative no.s are = -25, -24, -23, -22 and -21. 

~ Hope this helps :) 

So here's the problem figure 

From figure, 

\(tan 25 = {EF \over DE}\)

\(0.47={9 \over DE}\)

\(DE=19.15\)

∴ Length of DE is 19.15 cm.

Thanks Phill!! 

Lets assume there are \(x\) games in their collection. 

So, now we have 

Blake found = \({3x \over 4}\) video games 

Andy found = \({3x \over 4}-18\) video games 

Back to the question... total no. of video games are \(x\)

∴ \({3x \over 4}+{3x \over 4} -18=x\)

   \({6x \over 4}-x=18\)

   \(2x=72\)

    \(x=36\)

Hence, there are 36 video games in their collection. 

It is given that 

\(f(x)=x^2-2x+5\)

\(g(x)=x+8\)

∴ \(g(5) = 5 + 8 \)

            \(= 13\)

\(f(g(5)) = f(13) \)         

              \(= 169 - 2(13) + 5             \)

               \(=148\)

Now, 

\(f(5) = 25 - 10 +5\)

         \(=20\)

\(g(f(5))=g(20)\)

               \(= 20+8\)

               \(=28\)

∴ \(f(g(5)) - g(f(5)) = 148-28\)

                                    \(=120\)

This question is inappopriate as you've mixed up the names in the second line.. right question is- 

Jake and Don had the same amount of money at first. When Jake gave Don $36, 50% of the amount of money Don had left was the same as 75% of what Jake had left. How much did they have in total at first?

Here's the solution- 

Let us assume that Jake and Don had \($x\) at first.

Now, Jake gave $36 to Don 

∴ Amt. left with Jake = \(x-36\)

Amt. left with Don = \(x+36\)

According to ques.

50% of \((x+36)\) = 75% of \((x-36)\)

⇒\(50(x+36) = 75(x-36)\)

⇒\(50x+1800=75x-2700\)

⇒\(25x=3600\)

⇒\(x=144\)

∴ Initially Jake and Don had $144 with them.

Hope this helps :) 

Hey KF!  Kindly note that we aren't here to do your maths homework for you.. first attempt them yourself and then if you don't get 1 or 2 of them you may clear your doubts here. Instead reposting them several times won't be much use they're only gonna get ignored.