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maw [93]
3 years ago
9

#7 - #9 theres the pic

Mathematics
1 answer:
Ratling [72]3 years ago
6 0
#7
2,4,8,16
15,30,60,120

#9
1,2,3,
4,8,12
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It would be a right angle
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\bf (\stackrel{x_1}{5}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{7}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{7}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{5}}}\implies \cfrac{5}{-5}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{-1}(x-\stackrel{x_1}{5}) \\\\\\ y-2=-x+5\implies y = -x+7

6 0
3 years ago
3 letters without replacement 4 letters A B C D how many ways can this be done if the order of the choices matters
Veseljchak [2.6K]

Answer:

Since the order of choice matters, we will permute the values.                       a bit more explanation for this:

If the order of choice did NOT matter, ABC and BCA will be counted as one since order of choice does NOT matter

Since order of choice does matter, ABC , BCA and CAB are all different possibilities for the arrangement of the same 3 letters

Since we have 3 slots:

___  ___ ___

Now, for the first slot. You can out either one if the 4 alphabets in the first slot since no slot has been used as of now

So:

_<u>4</u>_ ___ ___

**Keep in mind that the 4 is the possible number of values this slot can have**

Now that one slot has been used, one of the 4 alphabets has been used and since we are not allowed to repeat the same alphabets, we are left with  3 more alphabets

we can put any one of the 3 alphabets in this second slot, Hence:

_<u>4</u>_ <u>_3_</u> ___

Now that 2 of the 4 alphabets have been used, we are left with only 2 alphabets, so there are only 2 possible alphabets for slot 3

Therefore:

_<u>4</u>_ _<u>3</u>_ _<u>2</u>_

Now that we know the possible alphabets for all 3 slots, we will multiply them with each other to get the total possible number of 3 - alphabet words we can make with 4 alphabets

Total possible words = 4 * 3 * 2

Total possible words = 24

We could've used the formula for Permutation as well

8 0
2 years ago
Can someone please help!!!! Will mark brainliest!
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b

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