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Mazyrski [523]
2 years ago
14

PLEASE HELP!!!!!! ILL GIVE BRAINLIEST *EXTRA 40 POINTS** !! DONT SKIP :((

Mathematics
2 answers:
elena-14-01-66 [18.8K]2 years ago
8 0

Answer:

yes

Step-by-step explanation:

I think sorry if I am wrong

katrin2010 [14]2 years ago
3 0
Answer
No
Explanation

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If A={4,7,10,13,17} and B={3,5,7,9}, then which of the following statements is true
EastWind [94]

Answer:

good questing it is something

Step-by-step explanation:

1-1-2- --3--1234-3

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- answer is 2def not cap i need 5 poinnts i gues

6 0
2 years ago
The mean age of 5 people in a room is 28 years.
lina2011 [118]

Answer:

174

Step-by-step explanation:

You should look at the mean and do this:

5*28 so now you know all the ages which is 140 now you do 140/29 and you get 4.8....... so you  can see that the answer is 174

4 0
3 years ago
Random question: hows life?
kipiarov [429]

Answer:

Great! Sorta making me really tired, but altogether fine!!!

How about you sir/ma'am?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please help....!!! EASY POINTS !!
Alex73 [517]
You should scan it on this app:)
8 0
3 years ago
If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
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