x is case of almonds and y is case of walnuts.
Almonds are packaged 15 bags per case and walnuts are packaged 17 bags per case.
H-E-B orders no more than 200 bags of almonds and walnuts at a time.
So,
x + y < 200
where x and y refers to the number of bags
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H-E-B pays $24 per case of almonds and $27 per case of walnuts, but will not order more than $300 total at any one time.
But keep in mind that : Almonds are packaged 15 bags per case and walnuts are packaged 17 bags per case.
So,
24 * (x/15) + 27 * (y/17) < 300
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The constraints are:
x + y < 200
(24/15) x + (27/17) y < 300
So, the graph of the previous constraints is as following :
6/16 & 3/8. Hope it helped.
Answer:
![3x^{3}+15x^{2}+x+5](https://tex.z-dn.net/?f=3x%5E%7B3%7D%2B15x%5E%7B2%7D%2Bx%2B5)
Step-by-step explanation:
We are told that the area of the bag can be represented by the function
and as the bag is raised up its height can be represented by the function
.
Since we know that we can find volume of cuboid by multiplying base area to its height. We are given area and height of bag as functions. Now we will find volume of the bag by multiplying these functions.
![\text{Volume of collapsible bag}=(3x^{2}+1)*(x +5)](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20collapsible%20bag%7D%3D%283x%5E%7B2%7D%2B1%29%2A%28x%20%2B5%29)
After using distributive property we will get,
![\text{Volume of collapsible bag}=3x^{2}(x+5)+1(x+5)](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20collapsible%20bag%7D%3D3x%5E%7B2%7D%28x%2B5%29%2B1%28x%2B5%29)
![\text{Volume of collapsible bag}=3x^{3}+15x^{2}+x+5](https://tex.z-dn.net/?f=%5Ctext%7BVolume%20of%20collapsible%20bag%7D%3D3x%5E%7B3%7D%2B15x%5E%7B2%7D%2Bx%2B5)
Therefore, the collapsible bag can hold
grain.
Answer:
What expression?
Step-by-step explanation:
Step-by-step explanation:
-4a-5b-2c I think that is the best answer