so, let's keep in mind that
so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.
we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.
![\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%2By%2B3y%3D78%5C%5C%20x%2B4y%3D78%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%200.25x%2B0.4y%2B0.6%283y%29%3D35.1%5C%5C%200.25x%2B0.4y%3D1.8y%3D35.1%5C%5C%200.25x%2B2.2y%3D35.1%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20x%2B4y%3D78%5C%5C%5C%5C%200.25x%2B2.2y%3D35.1%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20x%2B4y%3D78%5Cimplies%20%5Cboxed%7Bx%7D%3D78-4y%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20substitution%20on%20the%202nd%20equation%7D%7D%7B0.25%5Cleft%28%20%5Cboxed%7B78-4y%7D%20%5Cright%29%2B2.2y%3D35.1%7D%5Cimplies%2019.5-y%2B2.2y%3D35.1)
![\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill](https://tex.z-dn.net/?f=%5Cbf%201.2y%3D15.6%5Cimplies%20y%3D%5Ccfrac%7B15.6%7D%7B1.2%7D%5Cimplies%20%5Cblacktriangleright%20y%3D13%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20x%3D78-4y%5Cimplies%20x%3D78-4%2813%29%5Cimplies%20%5Cblacktriangleright%20x%3D26%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20z%3D3y%5Cimplies%20z%3D3%2813%29%5Cimplies%20%5Cblacktriangleright%20z%3D39%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B25%5C%25%7D%7B26%7D%5Cqquad%20%5Cstackrel%7B40%5C%25%7D%7B13%7D%5Cqquad%20%5Cstackrel%7B60%5C%25%7D%7B39%7D~%5Chfill)
The composite function combines the palm tree and the seed functions
The composite function is t(d) = 60d + 20
<h3>How to determine the composite functions</h3>
The functions are given as:
Number of palm trees: t(s) = 3s + 20
Number of seeds: s(d) = 20d
The composite function that represents the number of palm trees Carlos can expect to grow over a certain number of days is represented as:
t(s(d))
This is calculated as:
t(s(d)) = 3s(d) + 20
Substitute s(d) = 20d
t(s(d)) = 3 * 20d + 20
Evaluate the product
t(s(d)) = 60d + 20
Rewrite as:
t(d) = 60d + 20
Hence, the composite function is t(d) = 60d + 20
Read more about composite functions at:
brainly.com/question/10687170
Step-by-step explanation:
x=y+9
substitute
3(y+9)+8y=-6
3y+27+8y= -6
13y = -6 -27
13y= -33
y = -33/13
x = -33/13 + 9
Answer:
(-♾, 5) (5, ♾)
Step-by-step explanation:
Answer:
The new can can hold upto volume of 476.69 cubic centimeters.
Step-by-step explanation:
Diameter of the can = d = 6 cm
Radius of the can = r = 0.5d = 0.5 × 6 cm = 3 cm
Height of the can = h = 12 cm
Company wants to increase the dimensions of their cans by a multiple of 1.12. So, the new dimension will be:
r'= 3 cm × 1.12 = 3.36 cm
h' = 12 cm × 1.12 =13.44 cm
Volume of the = can = V
Volume of a cylinder = 
The new can can hold upto volume of 476.69 cubic centimeters.
