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)
Answer:
He can make only one batch of cookies.
Step-by-step explanation:
Just by mental math you can say he can make maximum one batch of cookies.
Total sugar = 10 cups
sugar used for lemonade = 2 cups
leftover sugar = 8 cups
For 1 batch of cookie sugar needed = 11/2
which means 5 and 1/2 cups
And he's just left with 8 cups of sugar after lemonade
8 - 5 1/2 = 2 1/2 which is not enough for 2nd batch
That means the maximum number of cookies batches is 1.
Another way to solve it by inequality:
2+Il/2 X <u><</u> 10
11/2 X <u><</u> 8
11X <u><</u> 16
X <u><</u> 1 6/11
So he can make only one batch of cookies.
Answer:
proved that x=9
Step-by-step explanation:
A line that splits an angle into two equal angles. ("Bisect" means to divide into two equal parts)
so <CAB=<DAB
7x+2=5(x+4)
7x+2=5x+20
7x-5x=20-2
2x=18
x=9
Answer:
308.647
Step-by-step explanation:
Its searchable pls mark brainliest.
C=18*13
C=234
You just have to multiply
