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Svetach [21]
3 years ago
11

In how many ways can a teacher arrange 6 students in the front row of a classroom with a total of 20 students?

Mathematics
1 answer:
Luba_88 [7]3 years ago
5 0
The concept of the question above is Permutation where in we are to arrange to arrange the 6 students from which there are 20 students. This is the "permutation of 20 taken 6" or 20P6. The numerical value of the permutation is equal to 27907200. 
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7^x^2=9^x <br>how do I solve this using logs
Rina8888 [55]
Apply lg on both sides.
lg7^x^2 = lg9^x
Bring the power forward.
x^2lg7 = xlg9
x^2/x = lg9/lg7
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4 0
2 years ago
Plz help!! marking brainiest!! (btw jus to clarify, i dont have three questions, they're jus graphs)
Ulleksa [173]

Answer:

c

Step-by-step explanation:

8 0
3 years ago
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Zoe is 16 years old. Her brother Luke is 3 years more than half her age. Write a numerical expression for Luke’s age.
Alenkasestr [34]

; Simplified =

Let "Zoe" = z

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3 0
3 years ago
Let a, b, c and x elements in the group G. In each of the following solve for x in terms of a, b, and c.
alina1380 [7]

Answer:

The answer is x=a^{-1}cb^{-1}.

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element g\in G there exist another element, denoted by g^{-1} such that gg^{-1}=e, where e stands for the identity element of G.

So, given the equality axb=c we make a left multiplication by a^{-1} and we obtain:

a^{-1}axb =a^{-1}c.

But, a^{-1}axb = exb = xb. Hence, xb = a^{-1}c.

Now, in the equality xb = a^{-1}c we make a right multiplication by b^{-1}, and we obtain

xbb^{-1} = a^{-1}cb^{-1}.

Recall that bb^{-1}=e and xe=x. Therefore,

x=a^{-1}cb^{-1}.

6 0
3 years ago
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SIZIF [17.4K]
That would be simplified into

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7 0
3 years ago
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