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Anettt [7]
2 years ago
11

Which expression is equivalent to the expression shown below?

Mathematics
2 answers:
OlgaM077 [116]2 years ago
7 0

Answer:

5^{15}

Step-by-step explanation:

(5^3)^{9} \div 5^{12}=5^{3 \times 9} \div 5^{12}=5^{27} \div 5^{12}=5^{27-12}=5^{15}

Georgia [21]2 years ago
5 0
5 to the power of 15

This is because you will need to times the powers by the brackets e.g 5x9 which is 45, then you need to takeaway 12 from that answer to get 15. You keep the 5 the same xx

Hope this helps!!
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In triangle ABC, m∠A = 36°, m∠B = 84°, and m∠C = 60°. The side lengths, in order from greatest to least, are , , .
katen-ka-za [31]

In the triangle ABC, the side lengths, in order from the greatest to the least, are : AC > AB > BC.

We are given a triangle. The vertices of the triangle are A, B, and C. The measures of the angles ∠A, ∠B, and ∠C are 36°, 84°, and 60°, respectively. We need to arrange the side lengths in order from the greatest to the least.

The side lengths are proportional to their opposing angles in a triangle. It means that the side opposite the largest angle is the largest side, and vice versa. The angles arranged in descending order are : 84° > 60° > 36°. The angles arranged in descending order according to the vertices are : B > C > A. The order of the lengths of the opposite sides must be the same.

Hence, the side lengths, in order from the greatest to the least, are : AC > AB > BC.

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5 0
1 year ago
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In e
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The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let \%R_{A} be the quantity of nonzero elements in R_{A}.

Let C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right).

Parity demands that \%R_{A} and\%R_{B} must equal 2 or 4.

Case 1: \%R_{A}=4 and \%R_B=4. There are \left(\begin{array}{l}3\\ 2\end{array}\right)=3 ways to put 2-1's in R_A, so there are 3 ways.

Case 2: \%R_{A}=2 and \%R_B=4. There are 3 ways to position the -1 in R_A, 2 ways to put the remaining -1 in R_B (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of \%R_{A}=4,\%R_{B}=2 for a complete of 24 ways.

Case 3: \%R_A=\%R_B=2. There are 3 ways to put the -1 in R_{A}. Now, there are two cases on what happens next.

  • The 1 in R_B goes directly under the -1 inR_A. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in R_C andR_D. (Either the 1 comes first inR_C or the 1 comes first in R_D.)
  • The 1 in R_B doesn't go directly under the -1 in R_A. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

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