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\(If \quad 625^x=64\qquad then\;find\qquad 125^x\\~\\ 625=5^4\\ 125=5^3\\ 64=4^3=2^6\\~\\ 625^x=64\\ 5^{3x\cdot \frac{4}{3}}=64\\ (125^x)^{\frac{4}{3}}=4^3\\ (125^x)=4^{3 \cdot \frac{3}{4}}\\ 125^x=4^{\frac{9}{4}}\\ 125^x=4^2*4^{1/4}\\ 125^x=4^2*2^{1/2}\\ 125^x=16\sqrt2 \)

NOTE:  I could easily have made a careless error.

LaTex

If \quad 625^x=64\qquad then\;find\qquad 125^x\\~\\
625=5^4\\
125=5^3\\
64=4^3=2^6\\~\\
625^x=64\\
5^{3x\cdot \frac{4}{3}}=64\\
(125^x)^{\frac{4}{3}}=4^3\\
(125^x)=4^{3 \cdot \frac{3}{4}}\\
125^x=4^{\frac{9}{4}}\\
125^x=4^2*4^{1/4}\\
125^x=4^2*2^{1/2}\\
125^x=16\sqrt2

8/125 ^x = 5/2       LOG both sides to get 

x LOG (8/125 ) = LOG (5/2) 

x =LOG(5/2) /  LOG (8/125) 

x =  - 1/3  

(8/125) ^-1/3  =  5/2 

f = fish #

5/8 f = shell number 

(5/8f -50) / (f-8)   = 1/4        <====  solve for f ....cross multiply to get:

  4 ( 5/8f-50) = f-8 

20/8f - 200 = f-8 

12/8 f = 192 

f = 128 

then ORIGINAL shells were   5/8 (128) = 80 

   FINAL shells = 80 - 50 = 30 shells in the end 

Hello turtlenob!

If there are 7 out of every 12 people who are bored by cable news, this means that 1 - 7/12 = 5/12 people who are not bored by cable news. If we call the total amount of people at the convention, x. Then 5/12 * x = 280. We get x = 672. Now we multiply 672 * 7/12 = 392 people who are bored by cable news.

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vockey

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I reworked the solution with two cases, and it is actually 336. 

We can have \(\begin{matrix} 6 & 4 & 1 \\ 2 & 1 & 1 \\ \end{matrix}\) for when the third seat is sibling of the first seat. There are 6 choices for the first seat. For the second seat there are 5 people left - 1 sibling of first seat = 4. Then, the third seat will be the sibling of the first seat, so 1. This leaves a pair of siblings and one person for the second row, so the only order that will work is \(\begin{matrix} s & p & s \end{matrix}\). This gives two choices from the two siblings for the second row.

Then, we will have \(\begin{matrix} 6 & 4 & 2 \\ 3 & 2 & 1 \end{matrix}\) for when the third seat is not the sibling of the first seat. 6 choices for first seat. 5 - 1 = 4 for the second seat. 4 people left -  1 sibling of first - 1 sibling of second = 2. The second row can have any order since there are no pairs of siblings left, so it is 3!.

\(6 \cdot 4 \cdot 1 \cdot 2 \cdot 1 \cdot 1 = 48\) and \(6 \cdot 4 \cdot 2 \cdot 3 \cdot 2 \cdot 1 = 288\). Summing both cases yields \(\fbox{336}\).

This is how I solved it. (Please tell me if it's correct.)

There are two possible arrangements of the siblings. First, there can be two pairs of siblings with each on the first or second row with one pair of siblings split between them:\( \begin{matrix} a & b & a \\ c & b & c \\ \end{matrix}\). Second, there is the arrangement where all pairs of siblings are split between rows:\( \begin{matrix} a & b & c\\ a & b & c \\ \end{matrix}\). We multiply \(2\) (the arrangements) by \(3!\) which is determined by which pairs of siblings go to \(a\), \(b\), and \(c\). Then, we multiply it by \(2^3\) which is determined by whether the first or second sibling sits in a chair for each pair of siblings. \(2(3!)(2^3)=\fbox { 96 }\)

First, we should find the coordinates for the x and y intercepts for the equation 3x - 6y = 12 + 8x - 3y.

x-intercept:

The x-intercept is the point at which the line passes through the x-axis. This means that y is always equal to zero. Thus, we can set the y values in the equation to zero. 3x - 6(0)= 12 + 8x - 3(0). Solving for x yields -12/5, so our x-intercept is at (-12/5, 0)

y-intercept:

The process is the same as for the x-intercept, except it is the x-value this time that is 0.  3(0) - 6y = 12 + 8(0) -3y. Solving for y yields a value of -4. Thus, our y-intercept is at (0, -4)

To find the distance between our ordered pairs of (-12/5, 0) and (0, -4), we use the distance formula: \(d = \sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2 }\) where our ordered pairs represent (x1, y1) and (x2, y2)

Plugging our variables into the formula gets our final answer of d = (4 sqrt(34))/5 ≈ 4.66476

Look at the function of g(x). How will changing f(x) to f((x^2 - 2)/(x + 1)) affect the domain?

The numerator of (x^2 - 2) only affects the range since all real numbers x can be plugged into it. From this, the domain is unaffected and stays at [-8,8].

Now, we must consider the denominator of (x + 1). Remember that the domain of a function is the set of possible values for x in the function. What is not a possible value here? Well, dividing by 0 would result in an undefined y value, so (x + 1) ≠ 0. Solving for x yields -1. This is the only x value that is not valid since all other real numbers x can be plugged into the denominator. Thus, the denominator constricts our new domain to [-8,1) U (1,8].

Now that we have the domain of f((x^2 - 2)/(x + 1)) which is equal to g(x), we can solve for the width of the interval of the domain of g(x). [-8 to 1) has a width of 9, and (1, 8] has a width of 7.  9+7=16, so 16 is our answer.

First, we observe the pattern of remainders when powers of 5 are divided by 7 (long division can help doing this by hand):

5^1 ≡ 5 (mod 7)
5^2 ≡ 4 (mod 7)
5^3 ≡ 6 (mod 7)
5^4 ≡ 2 (mod 7)
5^5 ≡ 3 (mod 7)
5^6 ≡ 1 (mod 7)
5^7 ≡ 5 (mod 7)
 

We see that the remainders repeat after every 6 powers. Since 207 is not a multiple of 6, we can express it like this: 207 = 34 * 6 + 3.

Therefore, we can write 5^207 ≡ 5^(34 * 6 + 3) ≡ (5^6)^34 * 5^3 ≡ 5^3 ≡ r (mod 7)

(the 5^(34*6) is negated due the cycle of remainders that is 6)

Finally, we wish to find r by calculating 5^3 mod 7. 5^3 yields 125. 125 / 7 = 17 remainder 6

Hence, the remainder when 5^207 is divided by 7 is 6.

@bingboy, I had the same basic idea as you, but I think that you missed that the question for all distinct paths that spelled ARCH. Also, I don't get how there were two ways to start with A. Correct me if I'm wrong, though. Here's my solution:

Starting with first A in the array, we can make a tree diagram that explores all paths that go A to R to C to H.
From A, we can reach the 3 Rs in 3 ways by moving up, down, or diagonally.
From each of the 3 paths to the Rs, there are 2 paths from 2 Rs to a C and 5 paths from the other R to a C.
Now, even though we can arrive at the same C through different paths, we must take into account that the problem wants all distinct paths. From the 8 total C paths that we arrived at, we can finish our word with any of the 7 Hs. We end up with 2+3+2+3+5+3+2+2+3 distinct paths at the end, which equals an answer of 25.
It's much easier to visualize this by drawing a tree which I've hopefully attached.
 

The probability that the interval I contains $2/3$ is $1/3.$ The probability that interval J also contains $2/3$ is $1/3.$ So, the probability that both I and J contain $2/3$ is $(1/3) \cdot (1/3) = 1/9$.

Let b = initial number of blue pens and r = initial number of red pens.

Form the following two equations and solve them for r and b:

\(\frac{b}{r}=\frac{4}{5}\\ \frac{b-280}{r-280}=\frac{9}{20}\)

Look at https://web2.0calc.com/questions/probability-problem_24#r2.  Use the same method, suitably modifying the numbers.

Construct a triangle ABC inside triangle XYZ with O also as its center. Any point inside ABC will be closer to O than a side length of XYZ, so we can find the probability by finding the ratio of [ABC] to [XYZ]. The distance from O to a side of ABC will be half the distance from O to a side of XYZ, so the altitude of ABC will be half of that of XYZ. Because XYZ and ABC are equilateral, the ratio of their side lengths will also be 1:2. The ratio of the area of ABC to the are of XYZ will be (1/2)^2, so the probability is 1/4.