Use BEDMAS
B-brackets
E-exponents
DM-multiply or divide (left to right)
AS-add subtract (left to right)
![45+22\div11\cdot\{(60-15)\cdot60]+25\}=45+2\cdot[(45\cdot60)+25] \\\\=45+2\cdot(2,700+25)=45+2\cdot2,725=45+5,450=5,495](https://tex.z-dn.net/?f=%2045%2B22%5Cdiv11%5Ccdot%5C%7B%2860-15%29%5Ccdot60%5D%2B25%5C%7D%3D45%2B2%5Ccdot%5B%2845%5Ccdot60%29%2B25%5D%20%5C%5C%5C%5C%3D45%2B2%5Ccdot%282%2C700%2B25%29%3D45%2B2%5Ccdot2%2C725%3D45%2B5%2C450%3D5%2C495%20)
Answer:
13 boxes
Step-by-step explanation:
To calculate the minimum number of boxes that can be had with that amount of paper, they should be kept in the largest box, that is, the 27 cubic foot box, therefore, we divide:
257/27 = 9.51
Which means that we can save the paper in 9 boxes of 27 cubic feet, now we will save the rest in the next smaller box, the 9 cubic foot box.
First let's calculate how many leftover paper:
257 - 27 * 9 = 14
Therefore, we can only use 1 box of 9 cubic feet:
14 - 9 = 5
For the remaining 5 cubic feet use 1 box of 3 cubic feet and 2 boxes of 1 cubic foot.
Therefore, in total it would be:
9 of 27 ft ^ 3 + 1 of 9 ft ^ 3 + 1 of 3 ft ^ 3 + 2 of 1 ft ^ 3 = 13
a total of 13 boxes would be the minimum
Answer:
Step-by-step explanation:
subtract 
from both sides
plz mark me brainliest. :)
X=27 I think this is correct
