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S_A_V [24]
2 years ago
11

Need help ASAP !!!

e=" \sqrt{2563 \times 2563} " alt=" \sqrt{2563 \times 2563} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
Vesnalui [34]2 years ago
5 0

Answer:

your answer will be -,+2563

Step-by-step explanation:

schepotkina [342]2 years ago
4 0

Answer:

√{2563×2563}=√2563²=±2563

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a study shows that 70 out of 200 college students read the newspaper. Marysville college has 12600 students. based on the study
Leviafan [203]
12600(7/20)=4410 students
3 0
3 years ago
Read 2 more answers
PLS HELP <br>find the unknown side of the triangle below (round to the nearest tenth) <br>​
Nonamiya [84]
22.9 is the closest but not exact. It was probably rounded off.

I used the formula: leg^2+leg^2=hypotenuse^2

(You can use that formula for any right triangle.)
7 0
3 years ago
3 − m = −2 (m + 6); m = −15 Yes or No
Nimfa-mama [501]

Answer:

3 − m = −2 (m + 6); m = −15    Yes

5 (p + 3) = −35; p = −4            No

8q = 3 (10 + q); q = 6             Yes


Step-by-step explanation:

3 − m = −2 (m + 6)

3 - m = -2m - 12

m = -15

Answer is Yes


5 (p + 3) = −35

5p + 15 = -35

5p = - 50

p = -10

Answer is  No


8q = 3 (10 + q)

8q = 30 + 3q

5q = 30

q = 6

Answer is  Yes

5 0
3 years ago
Help pls I will give brainliest!
oee [108]

Answer:

We are missing 210

Step-by-step explanation:

3*4 = 12

3*70 =210

3*300=900

We are missing 210

8 0
2 years ago
Read 2 more answers
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
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