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Yuliya22 [10]
3 years ago
9

Who knows that answer

Mathematics
1 answer:
nadezda [96]3 years ago
7 0
Choose a random number , square it , then add 20 
Ex) 2
          2^{2} = 4
          20+4=24

Just keep doing that till all the squares are filled 

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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
3/12 = y/8 solve for y
PIT_PIT [208]
2 is the answer for y
6 0
4 years ago
Evaluate 2x^2+5x^2-5x^3 when x = 2
Margaret [11]

Answer:


question is 2x^2+5x^2-5X^3   if the value of x=2 then we put the value of x in equation= 2(2)^2+5(2)^2-5(2)^3   =2*4+5*4-5*8    = 8+20-40   =28-40   =-12 ans


5 0
3 years ago
PLEASE HELP FIRST CORRECT ANSWER GETS BRAINLIEST AND LOTS OF POINTS
kykrilka [37]

Answer: y= -4

Step-by-step explanation:

7 0
3 years ago
Which point is on the graph of f(x)=2x= 2.5%?
katovenus [111]

Answer: The parabola has its concavity downwards, so we need a function in the model:

With a negative value of 'a'

The vertex is (0,0), so we have that:

The x-coordinate of the vertex is given by the equation:

So we have a function in the model:

With a < 0

The only option with this format is B:

Step-by-step explanation:

4 0
3 years ago
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