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AURORKA [14]
4 years ago
11

6X to the -8th times negative 3X to the -3

Mathematics
1 answer:
Westkost [7]4 years ago
4 0
Im pretty sure it's -18x^11
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I need an explanation
kap26 [50]

Answer:

equilateral x=20

Step-by-step explanation:

3x=60

/3 /3

x=20

all angles of equilateral triangles equals 60

7 0
3 years ago
Read 2 more answers
Joy ran 13 miles in 3 1/4 hours. what was her average rate ???
kotykmax [81]
13 miles/ (3 1/4 hours)= 4 miles/hour.

Her average rate is 4 miles/hour~
7 0
3 years ago
-2x + 3 = -11<br> X =<br> Helpppp
Fiesta28 [93]

Answer:

7

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Can someone tell me how to do this ?
ElenaW [278]

i am not 100% sure but...

(4x+3)=47

<u>    - 3    - 3 </u>

     4x=44

44/4= 11

so x = 11

7 0
3 years ago
How many sets of three integers between 1 and 20 are possible if no two consecutive integers are to be a set?
erastovalidia [21]
There are \dbinom{20}3=1140 total possible ways to pick any three integers from the set.

Of the total, there are 18 consisting of consecutive triplets (\{1,2,3\},\{2,3,4\},\ldots,\{18,19,20\}).

Now, of the total, suppose you fix two integers to be consecutive. There would be 19 possible pairs (\{1,2\},\{2,3\},\ldots,\{19,20\}), and for each pair 18 possible choices for the third integer (for instance, \{1,2\} can be taken with 3, 4, ..., 20), to a total of 19\times18=342. To avoid double-counting (e.g. \{1,2\} can't go with 3; \{2,3\} can't go with 1 or 4), we subtract 1 from the extreme pairs \{1,2\} and \{19,20\} (twice), and 2 from the rest (17 times).

So, the number of triplets that don't consist of pairwise consecutive integers is

1140-(18+342-(2\times1+17\times2))=816

I don't know how useful this would be to you, but I've verified the count in Mathematica:

In[8]:= DeleteCases[Subsets[Range[1, 20], {3}], x_ /; x[[2]] == x[[1]] + 1 || x[[3]] == x[[2]] + 1] // Length
Out[8]= 816
6 0
3 years ago
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