Would you rather lose the ability to cry or cry every day for 20 minutes randomly?

Mathematics

2 answers:

AlladinOne [14]2 years ago

8 0

Answer:

Lose the ability to cry because i dont cry that much

Step-by-step explanation:

Dennis_Churaev [7]2 years ago

3 0

Answer:

loose the ability to cry l m a o

Step-by-step explanation:

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Tyger is trying to eat less sugar in his diet.He reduces his sugar intake from 15 spoons a day to 9 1/2 spoons a day. What is th

seraphim [82]

Answer:the percentage decrease is 36.67%

Step-by-step explanation:

Initial amount of sugar in his diet is 15 spoons in a day. Currently, the amount of sugar in his diet is 9 1/2 spoons in a day. Converting to decimal, it becomes 9.5 spoon in a day. The amount by which he reduced his sugar intake is 15 - 9.5 = 5.5 spoons.

Percentage decrease would be expressed as

Amount by which the intake decreased/ initial intake × 100

It becomes

5.5/15 × 100 = 36.67%

8 0

2 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.