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Hi Straight and guest.

It makes it ieasier if you think of each die being a different colour.

1*2*6=12         3!= 6 ways

1*3*4=12               6 ways

2*2*3=12              3 ways

that is it I think

6+6+3 = 15 ways

6*6*6 = 216

so the prob will be    15 / 215  =     5 / 72

If he used 1/6 of his salary to pay the household bills, how much was his total salary?

Hello Guest!

What is used in the household accounts in the ratio 3: 2 please?

  !

U did the opposite way

\(x = 1 + \cfrac{\sqrt{3}}{1 + \cfrac{\sqrt{3}}{1 + \dotsb}}. Find \frac{1}{(x + 1)(x - 2)}. \)When your answer is in the form \(\frac{A + \sqrt{B}}{C}\), where A, B, and C are integers, and B is not divisible by the square of a prime,

\(x = 1 + \cfrac{\sqrt{3}}{1 + \cfrac{\sqrt{3}}{1 + \dotsb}}\) When we look at this fraction, we see that the denomenator of the fraction is also equal to x, because it's going on forever. Therefore, we can just change this equation to \(x=1+\sqrt{3}/x\). 

Now, you can just solve for x and plug it into 1/(x+1)(x-2) and get your answer. Hope this helps!

(correct me if i am incorrect)

We have f(x) = ax^r

If f(2) = 1, that means that a*2^r = 1

If f(128) = 5, that means that a*128^r = 5

We currently have two equations with two variables each.

\(a*2^r = 1\)

\(a*128^r=5\)

now, it's just a system of equations, solving for a and r. Can you take it from here?

INversely proportional

j = c/k     or

jk = c       c = constant

42(56) = 2352

when k = 98

j (98) = 2352

j = 2352/98 = 24         just as ahz001 found !

Originally, there were 300 students in honor roll. Now we increased boys by 24% and girls by 6%.

Boys: 150 * 0.24 + 150 = 186 

Girls: 150 * 0.06 + 150 = 159

Adding those up, this year we have 186 + 159 = 345 students in honor roll this year.

Now, we can form an equation, using our previous ones as an example.

x = total percent increase

300 * x + 300 = 345

300x = 45

x = 0.15 = 15%

(correct me if i am incorrect)

If they're inversely proportional, then \(j*k=j*k\) even when j and k are different. In this case, \(42*56=j*98\). 

We would have \(j= (48*56)/98\)

Solving this, we have \(j = 24\) 

(correct me if i am incorrect)

I have not done this question but this is how I would try tackling it.

work out formula for the distance between (0,0)  and  (  x, (x^2-18)/sqrt2   )

Then use calculus to minimize it.

Before hertrip to Hong Kong, Oga exchanged C$500 for HK$3287. About how many Hong Kongdollars did Oga receive for each Canadian dollar?

You want to know     Hong Kong dollars  PER  Canadian dollar

So you take the Hong Kong amount and DIVIDE it by the canadian amount.

Please put your latex into a latex box (on the ribon) so that it displays properly.

The answer is S, P, Q, R.

6.574???

I'm not sure, but there are 46 Spanish students and 51 Chinese students.

97 students speak at least one language.

27 of them are only Spanish. ONLY

Means (46 - 27 =) 19 students speak two languages (Spanish and Chinese).

But it could be more because some of the Chinese students can maybe speak two languages, there is no information about this.

x is the total of students that speak two languages.

So \(19 \leq x \leq 70\) . 

a = 10 and b = 2    Oh yeah??? 

Problem says:  "If you multiply 'a' by the sum of 'a' and 'b', you get 30."  

a * (a + b) = 30 

10 * (10 + 2) = 30  NOT       No wonder you don't show your work. 

318 - 27 = 291.

The answer is 291 pieces of candy.

That's wrong, Guest. The solution is

a = 1

b = 29 .

But at least you tried.

If the order will matter, then (I think) the probability would be 5/72.

If the order does not matter, then (I think) the probability would be (still) 1/72.

The answer is wrong, I see. Let me solve it again.

I solved this problem before, and the answer is 23.

If the order will matter, then (I think) the probability would be 1/12.

If the order does not matter, then (I think) the probability would be 1/72.

a = 10 and b = 2.

365  *   5.48  *  1.09 =  $...................

Answer: 10

Explanation:

Simplifying that expression turns it into \(x^2+(k-5)x+36\). The 36 at the end (and the fact that the coefficent of x^2 is 1) shows us that there are only two possible options: (x+6)^2 and (x-6)^2. Expanding the first gives x^2 + 12x +36, while the second gives x^2-12x+36.

Set up an equation for the first one:

k-5=12

k=17

Set up an equation for the second one:

k-5=-12

k=-7

-7 + 17 = 10.