Twenty boxes of paper weigh 520 pounds.What is the ratio of boxes to pounds?

Mathematics

1 answer:

Akimi4 [234]2 years ago

8 0

the ratio of boxes to pounds is 20:520

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What is 2(3t-1)-5(t-1)

givi [52]

2(3t - 1) - 5(t - 1)
2(3t) - 2(1) - 5(t) + 5(1)
6t - 2 - 5t + 5
6t - 5t - 2 + 5
t + 3

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3 years ago

Can someone please help me with this

k0ka [10]

Answer: $3x^6+x^5-x^4-6x^3+x^2+3x-1$

Step-by-step explanation:

(3x^2+x-1)(x^4-2x+1)

=3x^6-6x^3+3x^2+x^5-2x^2+x-x^4+2x-1

=3x^6+x^5-x^4-6x^3+x^2+3x-1

Hope this helps!! :)

Please let me know if you have any question

5 0

2 years ago

Read 2 more answers

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.