Estimate 183 to the nearest integer.

Mathematics

1 answer:

wariber [46]2 years ago

6 0

Answer:

Step-by-step explanation:

1,83 x 10^2

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write an enaquality that expresses the reason the lengths 5 feet , 10 feet and 20 feet could not make a triangle. explain how th

NikAS [45]

For any triangle, the 2 shortest sides added together have to be longer than the longest sides.

short side (10) + short side (5) needs to be greater than (>) longest side (20)

10 + 5 > 20

15 > 20

This inequality is incorrect, so the lengths cannot make a triangle.

Hope this helps :)

7 0

3 years ago

7(2 + 3) − 4 · 2 2 the second two is up in the air fyi

AfilCa [17]

Answer:

Step-by-step explanation:

7(2 + 3) − 4 · 2² ← like that right so

parenthesis first add the numbers up↓

7(5) - 4 · 2²

write number in exponential form↓

7(5) - 2²· 2²

multiply↓

35 - 2² · 2²

calculate the product↓

35 - 2⁴

evaluate the power↓

35-16

then subtract↓

7 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.