Answer:
The answer is "
"
Step-by-step explanation:
Please find the graph file of the question in the attachment.
Its first step is the lightest and heaviest evil. 
is the lightest bag, and 
is the heaviest bag. To remove it now, it is not permissible to remove it as a mixed fraction, therefore convert each fraction into an improper fraction by multiplying the whole integer to the numerator, then adding a numerator to this amount, multiplied by the numerator. Follow those steps to find out is the wrong part.
The process is the same for 
You now deduce the numerator but just not the negative, to deduct both. You subtract, as the question is raised far as
is involved. The last stage is that this is transformed into blended families, dividing its count by the denominator, 6 divided by 4 is identical to 1.5. 1 is a full amount so you don't modify it, but you do need to change the 5 decimals to 
and the last step is to reduce.
It depends on how many numbers there are.
The answer is (-1,7)
I got this answer because to solve a system, you find the point on the graph where both lines intersect. And that point in this graph is (-1,7). Hope this helps!:)))))
"<span>y + 5 ≥ 4" is the one inequality among the following choices that has been shown. The correct option among all the options that are given in the question is the third option or option "C". The other choices are incorrect and can be easily avoided. I hope that this is the answer that has actually come to your desired help.</span>
let's recall the remainder theorem.
we know that (x-1) is a factor, that means x -1 = 0 or x = 1.
since we know that (x-1) is a factor, then dividing the polynomial by it will give us a remainder of 0, which correlates with saying that f(1) = 0, in this case, so we can simply plug in "1" as the argument, knowing it gives 0.
![f(x)=3x^3+kx-11\\\\[-0.35em]~\dotfill\\\\\stackrel{0}{f(1)}=3(1)^3+k(1)-11\implies \stackrel{f(1)}{0}=3+k-11\implies 0=-8+k\implies 8=k](https://tex.z-dn.net/?f=f%28x%29%3D3x%5E3%2Bkx-11%5C%5C%5C%5C%5B-0.35em%5D~%5Cdotfill%5C%5C%5C%5C%5Cstackrel%7B0%7D%7Bf%281%29%7D%3D3%281%29%5E3%2Bk%281%29-11%5Cimplies%20%5Cstackrel%7Bf%281%29%7D%7B0%7D%3D3%2Bk-11%5Cimplies%200%3D-8%2Bk%5Cimplies%208%3Dk)
