A rectangular prism is defined by three lengths.
We can find out how many unit cubes would be in a prism by multiplying these three lengths together--that's how we find our <em>volume</em>.
Similarly, we can come up with different ways to multiply together three different numbers and make 18.
Each combination would be a new rectangular prism, with one catch:
Order doesn't matter. A prism with lengths 2, 2, and 3 is the same as one with lengths 2, 3, and 2, so don't make that mistake.
To find each combination, keep splitting 18 in different ways.
If one of the ways we split it can also be split, we need to write out that, too.
Here are the possible combinations:
18 × 1 × 1, obviously
9 × 2 × 1. splitting off 2
6 × 3 × 1. splitting off 3
4 × 6 × 1. our next biggest we can take out is 6, which can also be split...
4 × 3 × 2. there's the split of 6 into 2 and 3
<em>(3 × 6 × 1 is a repeat.)</em>
3 × 3 × 2 is new, though
<em>(2 × 9 × 1 is a repeat...)
</em><em>(2 × 3 × 3 is a repeat...)
</em>(aaaand 1 × 1 × 18 is a repeat. let's count up our combinations.)
<em>
</em>
There are 6 possible ways to multiply numbers together and get 18...
So 6 possible rectangular prisms.
The answer would be A since in that case its a percentage and there can be multiple answers. In the other questions, there is a definitive answer for it.
Hope this helps :)
Let's say our first integer is "a".
how to get the next consecutive EVEN integer? well, just add or subtract 2 from it, therefore, the second consecutive integer will be "a + 2".
and the next after that, will then be (a + 2) + 2, or "a + 4".
so those are are 3 integers, a a + 2 a+4
notice that, from any even or odd integer, if you hop twice either forwards or backwards, you'll land on another even or odd integer respectively.
2 + 2 is 4, or 8 + 2 is 10 some even ones
3 + 2 is 5, or 13 + 2 is 15, some odd ones
![\bf \stackrel{\textit{3 times the first}}{3a}~~=~~\stackrel{\textit{26 less than twice the sum of the others}}{2[~(a+2)+(a+4)~]~~~-26} \\\\\\ 3a=2[~2a+6~]-26\implies 3a=4a+12-26\implies 3a=4a-14 \\\\\\ 0=a-14\implies 14=a](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7B3%20times%20the%20first%7D%7D%7B3a%7D~~%3D~~%5Cstackrel%7B%5Ctextit%7B26%20less%20than%20twice%20the%20sum%20of%20the%20others%7D%7D%7B2%5B~%28a%2B2%29%2B%28a%2B4%29~%5D~~~-26%7D%0A%5C%5C%5C%5C%5C%5C%0A3a%3D2%5B~2a%2B6~%5D-26%5Cimplies%203a%3D4a%2B12-26%5Cimplies%203a%3D4a-14%0A%5C%5C%5C%5C%5C%5C%0A0%3Da-14%5Cimplies%2014%3Da)
what are the other two consecutive integers? well, a + 2 and a + 4.
A) The situation represents an arithmetic sequence because the successive y-values have a common difference of 210.
F(1) = 240 +210
F(2) = 240 +2(210)
F(3) = 240+3(210)
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F(x)= 240 +210x.
Learn more about Sequence:
brainly.com/question/12246947
#SPJ4
Answer:
5.5193237 times 10 to the power of -10
Step-by-step explanation:
To solve you just need to divided the area by the length or width the question has provided. Please note I am in 7th grade.