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See the answer here:  https://web2.0calc.com/questions/fractions_76942

Based on the information given, we can represent the grid as follows:

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A-B-C-D-E-F-G-H-I

Since each step must be in a rightward direction, the only possible paths are those that go from left to right across the grid. We can count the number of such paths by counting the number of ways to choose the steps from the available options.

Starting at A, we have only one option: to step to B. From B, we have two options: we can either step to C or we can step directly to D. From each subsequent node, we have two options: we can either stay on the same row and move to the next column, or we can move diagonally up to the next row and then move to the next column. Using this method, we can count the total number of paths from A to I as follows:

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A -> B -> C -> D -> E -> F -> G -> H -> I 1 2 2 4 4 8 8 16 16

Therefore, there are 16 possible paths from A to I, if each step must be in a rightward direction.

H/J ==5/3...................................(1)
0.20H ==round
0.60J ==round
R==Number of round biscuits transferred.
0.20H + R==0.60J - R..............(2)
S==Number of square biscuits transferred.
0.80H==square
0.40J==square
0.80H - S ==0.40J + S.............(3)
R  +  S ==162............................(4), solve for H, J, R, S

Using the above 4 equations with substitutions and eliminations we get:
H==450 - original number of biscuits
J== 270 - original number of biscuits
R==36 -  round biscuits transferred from J to H
S ==126 - square biscuits transferred from H to J

450 + 36 - 126 ==360 biscuits in H after the transfer.
270 - 36 + 126 ==360 biscuits in J after the transfer.
H - J ==450 - 270 ==180 - extra number of biscuits that were in container H than container J at first.

Hmm, exact same diagram in both posts, looks like cheaters are in the same class

Let's start by assigning variables to represent the unknown quantities in the problem:

Let x be the number of 10-cent coins in the basket.

Then, the number of 20-cent coins in the basket is 2x, since it's given that there are 100% more 20-cent coins than 10-cent coins.

After 13 twenty-cent coins are exchanged for 10-cent coins, there are 2x-13 twenty-cent coins left in the basket, and the number of 10-cent coins in the basket is x+13*(2/1) = x+26.
According to the problem, after the exchange, the number of 20-cent coins left in the basket is 5/8 less than the number of 10-cent coins. Using this information, we can write the following equation:

2x - 13 = (3/8)*(x + 26)

Multiplying both sides by 8 to eliminate the fraction, we get:

16x - 104 = 3x + 78

Solving for x, we get:

13x = 182

x = 14 - Number of original 10-cent coins

Then, the number of 20-cent coins in the basket before the exchange was 2x = 28.

The total value of the coins in the basket before the exchange was (14 x 10) + (28 x 20) = 700 cents or $7.00

Therefore, there is $7.00 worth of coins in the basket.

Maybe this user didn't see the other question and needed some help. You didn't need to call them an unkind word, even if it was censored.

Can you show the grid?

Hi! Thanks so much for the quick response!!

I don't really undestand how the number of the possibilities for no X's in the first four letters could be bigger than the total possibilities?

Please explain because that will help so much!

Thanks so much anyways :)

I truly agree about the second point. However, scientifically it makes sense that the chicken came first. This is because the egg cannot develop without the chicken, but the chicken did develop without the egg (in my previous answer).

Harold has 48 marbles at first.

There are: 12! / [4!4!4!] = 34,650 - total number of 12-digit permutations.

Since there are no X's in the first 4 letters, the first 4 letters of the word must consist of only Y's and Z's. There are 8 such letters, so we can choose any 4 of them to be Y's and the remaining 4 to be Z's. The number of ways to do this is:

C(8,4) = 70

Once we have selected the first 4 letters, we can arrange the Y's and Z's in any order in the remaining 8 positions. There are C(8,4) ways to choose the positions of the Y's, and the Z's will occupy the remaining positions. Therefore, the number of such 12-letter words is:

70 * C(8,4) = 70 * 70 = 4,900

So there are 4,900 such words that have no X's in the first 4 letters.

I understand that we have to split the hexagon into different parts, and I know how to find the area of the right triangles and the squares, but how do we find the area of the non-right triangles, and what would the final answer be?

their iq

Look at this cheater, demanding us to give them an "anser". Probably failed first grade math. I'll give you an anser to you iq, and its very low

I think that the area of all of the triangles should be the same regardless of a and b.

wow, so demanding

don't expect us to give you any more effort than you put into writing this question, and if some hapless simp does, go toss yourself off a cliff because you are more likely to survive that

Hola Perracita, why don't you do it yourself and actually learn?

1. First, we notice that triangle OCP is a right triangle with hypotenuse OP and angle OCP = 90 degrees. Therefore, we can use the Pythagorean theorem to find CP:

CP^2 = OP^2 - OC^2 = 20^2 - r^2

Next, we notice that triangle CPQ is similar to triangle COA (by angle-angle similarity), so we have:

CP/QP = CO/OA

Substituting CP from the previous equation and simplifying, we get:

(20^2 - r^2)/QP = r/(2r)

Simplifying this equation and solving for QP, we get:

QP = 40r/(20^2 - r^2)

We also know that PQ = 7, so we can set up another equation:

PQ^2 + QO^2 = PO^2

Substituting PQ and QP from the previous equations and simplifying, we get:

7^2 + (40r/(20^2 - r^2) - r)^2 = 20^2

Expanding and simplifying this equation, we get a quadratic equation in r^2:

r^4 - 400r^2 + 9600 = 0

Using the quadratic formula, we get:

r^2 = (400 ± √(400^2 - 4*9600))/2 = 200 ± 20√5

Since r is the radius of the circle, we know that r > 0, so we take the positive square root:

r^2 = 200 + 20√5

Therefore, the answer is 200 + 20√5.

Hola Perracita. You are a non-contributing member to our society.