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Hello Guest,

\(\frac{m}{m^2+5m-6}+\frac{2}{m+6}=\frac{1}{m-1}\) ,

\(\frac{m}{m^2+5m-6}+\frac{2}{m+6}-\frac{1}{m-1}=0\) ,

\(\frac{m}{m^2+6m-m-6}+\frac{2}{m+6}-\frac{1}{m-1}=0\) ,

\(\frac{m}{m*(m+6)-m-6}+\frac{2}{m+6}-\frac{1}{m-1}=0\) ,

\(\frac{m}{m*(m+6)-(m+6)}+\frac{2}{m+6}-\frac{1}{m-1}=0\) ,

\(\frac{m}{(m+6)*(m-1)}+\frac{2}{m+6}-\frac{1}{m-1}=0\) ,

\(\frac{m+2(m-1)-(m+6)}{(m+6)*(m-1)}=0\) ,

\(\frac{m+2(m-1)-m-6}{(m+6)*(m-1)}=0\) ,

\(\frac{2(m-1)-6}{(m+6)*(m-1)}=0\) ,

\(\frac{2m-2-6}{(m+6)*(m-1)}=0\) ,

\(\frac{2m-8}{(m+6)*(m-1)}=0\) ,

\(2m-8=0\) ,

\(2m=8\) ,

\(m=4\) ,

so the solution is \(m=4\) .

Straight

Hello Guest,

it's me again, Straight, The answer is 9996. In the Orange party there are 11424 votes and then "which was seven eightshs of the Blue party vote that means \(\frac{7}{8}\) \( {of\ the\ Orange\ party}={Blue\ party}\). You know there are 11424 votes for the Orange party so: \(\frac{7}{8}\)\(*\ 11424\ votes\) and that's 9996 votes.

So I think my answer is correct.

Straight

Hello Guest,

(correction: Celia got 80% in her first three exams.)

Her reasoning is wrong because it would be right if she has written

one exam, this would be \(\frac{80+94}{2}\), which is 87%, but there are three

exams and not just one. So it equals: \(\frac{80+80+80+94}{4}\), which is 83.5%.

So her overall mark is no 87%, but it is 83.5%. It applies:

\(\frac{Sum\ of\ values }{Number\ of\ values}\) and that is the overall mark formula.

Straight

That is not correct, the answer is 50

8.8 hours...........

no the answer is incorrect 

hello Guest,

the result is -12.

Steps:

\(-2*6\)

\(-12\)

Straight

Yes, that's right,

and when the question is "What is the smallest three-digit multiple of 9 whose digits' sum is 18", it would be 189, I guess.

Straight

Hello Guest,

no, think again about what you are saying.

You will get the answer.

Straight

2022^2021+2019^2020      can you say the last digit

Only considering the last digits

2^1=2

2^2=4

2^3=8

2^4=*6

2^5=*2

2^6=*4

2^7=2^3=*8

2^8=2^4=*6

2^9=2^(5+4)=2^5=*2

so

I need to put  2021 =   1+ 4*505

So the last digit of  2022^2021 will be 2

-------------------------

Now look at powers of 9

9^1=9

9^2=81

9^3=729

9^4 = 6561

9^(1+2n)= **9

9^(2+2n)=**1

9^2020=9^(2*1010)   the last digit will be 1

---------------------

2022^2021+2019^2020  the last digit will be  2+1=3

You need to check what I have done.

Вам нужно проверить, что я сделал

Let

f(x) = 12 - 2x if x <= 4

f(x) = ax + b if x > 4

The function f has the property that f(f(x)) = x for all x. What is a + b?

I started by drawing the g part I know:

I can see from this that if x<4 then  f(x)>4

f(0)=12-0=12

f(12)=12a+b=0

12a+b=0   (1)

f(1)=12-2=10

f(10)=10a+b=1

10a+b=1 (2)

Solve (1) and (2)  simultaneously and you get a=-0.5  and b=6

I checked this answer and it worked fine.

Hello,

m and n can only be bigger than 4125 or 2816000,

\(m=15\) \(\frac{15!}{4125}=317011968\) and

 \(n=15\) \(\frac{15!}{2816000}=464373\) .

n is also 15.

Straight

It's \(2*(9+7)\) and that equals \(2*16\) and that is 32.

Hope this was hopeful.

Straight

The answer will be a multiple of 5 if one, or both, of the numbers ends in 0 or 5.

There are 90 numbers in total and 18 of them end in 0 or 5

The prob that the first one is a multiple of 5 and the second one is not is   \(\frac{18}{90}*\frac{72}{89}=\frac{72}{445}\)

The prob that the first one is not a multiple of 5 and the second one is will be the same    72/445

The prob that they are both multiples of 5 is   \(\frac{18}{90}*\frac{17}{89}=\frac{17}{445}\)

P(that the multiple will be a multiple of 5) = \(\frac{2*72+17}{445}=\frac{161}{445}\)  

You need to check for careless errors.

Given that 2^x=7, find the value of 4(x - 1).

Hello Guest!

\(2^x=7\\ x\cdot ln\ 2=ln\ 7\\ x=\dfrac{ln\ 7}{ln\ 2}\)

\(4(x - 1 )=4\cdot (\dfrac{ln\ 7}{ln\ 2}-1)\)

\( 4(x - 1)=7.229...\)

  !

x + xy = 250y

x - xy = 240y

Enter all possible values of x, separated by commas.

Hello Guest!

\(x + xy = 250y\\ x(1+y)=250y\\ x - xy = 240y\\ x(1-y)=240y \)

\(\frac{250y}{1+y}=\frac{240y}{1-y}\\ 250y-250y^2=240y+240y^2\\ 10y=490y^2\\ 0=y(490y-10)\)

\(\large y\in \{0,\dfrac{1}{49}\}\)

\(x=\dfrac{250y}{1+y}\)

\(x=\dfrac{250\cdot \dfrac{1}{49} }{1+\dfrac{1}{49}}=\dfrac{250}{49}\cdot \dfrac{49}{50}=5\)

\(\large x\in \{0,5\}\)

  !

Draw it up on a grid and then use Pythagoras's theorem to find the hypotenuse.

The times the answer by 0.25

The answer is in miles

Thanks Alan,

I was hoping you would look at this one  

Like so:

You might prefer to express this as \(\theta=...\) in which case I'll leave you to do the rearrangement.

Two real numbers are chosen at random between 0 and 2. What is the probability that the sum of their squares is no more than 1? Express your answer as a common fraction in terms of pi.

Sample space is represented by the area of the square, which is 4

The quadrant area includes all the favorable events.   x^2+y^2<= 1     ARea is    1/4 * pi *1^1 = pi/4

Probability of a favourable event is    pi/4 divided by 4     =    pi/16

That is not a hard one.  

It is best for you to work it out for yourself.   

Just think about +,  -,   multiply and divide.   But you don't need all of them.

Prove:

\((c^{2}e^{2}+d^{2}f^{2}) \leq (c^{2}+d^{2})(e^{2}+f^{2})\)

I have assumed that all pronumerals represent real numbers.

\(LHS=(c^{2}e^{2}+d^{2}f^{2}) \\ RHS= (c^{2}+d^{2})(e^{2}+f^{2})\\ RHS= c^{2}e^2+d^{2}f^2+c^2f^2+d^2f^2\\ RHS= LHS +c^2f^2+d^2f^2\\ since\;\;c^2\ge0,\quad f^2\ge0,\quad d^2\ge0 \quad and\quad f^2\ge0 \\ RHS= LHS +k\qquad \text{Where k is a real number }\ge0\\ LHS +k =RHS\\ LHS=RHS-k\\ LHS\le RHS\qquad QED \)

Similar triangles.