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3 years ago

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How many sets of four consecutive positive integers are there such that the product of the four integers is less than $100,\!000

$

1 answer:

SVEN [57.7K]3 years ago

4 0

Answer:

16,17,18,19

Step-by-step explanation:

one guy guessed. probably using a calculator

another guy took the 4th root of 100,000 and rounded down a number

(a third way started like this

n*(n+1)*(n+2)(n+3)=100000

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#82 will give brainliest to best answer!

Fofino [41]

Answer:

$\frac{6}{k-6}$

Step-by-step explanation:

First, we can factor all of the following equations to turn that weird, huge looking thing into $\frac{(k+6)(k-6)}{(k-6)(k-10)}$ ÷ $\frac{(k-6)^2}{k(k-6)}$ × $\frac{6(k-10)}{k(k + 6)}$ . We know that division is simply multiplication by the reciprocal, so that whole equation will turn into $\frac{(k+6)(k-6)}{(k-6)(k-10)}$ × $\frac{k(k-6)}{(k-6)^2}$ × $\frac{6(k-10)}{k(k+6)}$ . Now we can cancel out some values if they are both in the numerator and denominator, which will turn that still huge looking thing into $\frac{6}{k-6}$ which is our final answer, as it cannot be simplified further.

Hope this helped! :)

7 0

2 years ago

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What is this, x-8 all over 3=3

inysia [295]

You mean (x-8)/3 = 3 ...right?

You would multiply both sides by 3.
Then, add 8 to both sides.

x=17

5 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

9234 divided by 52 whats the answer

BlackZzzverrR [31]

The answer is 177.57 please mark me as brainliest since I'm first to answer, thanks,~Kashout kam

4 0

3 years ago

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If r = -18, s = 27, and t = -15, what is the value of r - s - t

7nadin3 [17]

-18- -27 = -45 -45- -15 = -30

8 0

3 years ago

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