P=1.15(20+0.5x)=23+0.575x
62.1=23+0.575x
0.575x=39.1
x=68
68 pages.
Answer:
![r=\frac{41}{14}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B41%7D%7B14%7D)
Step-by-step explanation:
![\frac{3r-2}{5} =\frac{5+2r}{8} \\\\8(3r-2)=5(5+2r)\\24r-16=25+10r\\24r-10r=25+16\\14r=41\\\\r=\frac{41}{14}](https://tex.z-dn.net/?f=%5Cfrac%7B3r-2%7D%7B5%7D%20%3D%5Cfrac%7B5%2B2r%7D%7B8%7D%20%5C%5C%5C%5C8%283r-2%29%3D5%285%2B2r%29%5C%5C24r-16%3D25%2B10r%5C%5C24r-10r%3D25%2B16%5C%5C14r%3D41%5C%5C%5C%5Cr%3D%5Cfrac%7B41%7D%7B14%7D)
7ysq2 + 7y
Simply add the coefficients of like terms. Addition is both associative and commutative.
Since we're looking for the rate of kilograms per minute, we can represent the rate as
![R \frac{kg}{min}](https://tex.z-dn.net/?f=R%20%5Cfrac%7Bkg%7D%7Bmin%7D)
. We just need to divide the kilograms crushed by the time it takes to crush them to obtain R:
Answer: 21
Step-by-step explanation:
Given: Number of mango guava and sapota are supplied 108,84 and 60 respectively.
Since, 108 = 2 x 2 x 3 x 3 x 3
84 = 2 x 2 x 3 x 7
60 = 2 x 2 x 3 x 5
HCF(108,84,60) = 2 x 2 x 3 [HCF=Highest common factors]
= 12
Greatest number of plants in each row = 12
Number of rows for mango = ![\dfrac{108}{12}=9](https://tex.z-dn.net/?f=%5Cdfrac%7B108%7D%7B12%7D%3D9)
Number of rows for guava = ![\dfrac{84}{12}=7](https://tex.z-dn.net/?f=%5Cdfrac%7B84%7D%7B12%7D%3D7)
Number of rows for sapota = ![\dfrac{60}{12}=5](https://tex.z-dn.net/?f=%5Cdfrac%7B60%7D%7B12%7D%3D5)
Total rows = 9+7+5= 21
Hence, the minimum number of rows in which they can be planted if in each row same numbers of saplings of the same type are to be planted = 21