![\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases} 343=7\cdot 7\cdot 7\\ \qquad 7^3\\ 36=6\cdot 6\\ \qquad 6^2\\ 256=4\cdot 4\cdot 4\cdot 4\\ \qquad 4^4 \end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}} \\\\\\ \sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4} \\\\\\ 7^2+6-4^3\implies 49+6-64\implies -9](https://tex.z-dn.net/?f=%5Cbf%20343%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B36%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-256%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cbegin%7Bcases%7D%0A343%3D7%5Ccdot%207%5Ccdot%207%5C%5C%0A%5Cqquad%207%5E3%5C%5C%0A36%3D6%5Ccdot%206%5C%5C%0A%5Cqquad%206%5E2%5C%5C%0A256%3D4%5Ccdot%204%5Ccdot%204%5Ccdot%204%5C%5C%0A%5Cqquad%204%5E4%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%20%287%5E3%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B%286%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-%284%5E4%29%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B%287%5E3%29%5E2%7D%2B%5Csqrt%5B2%5D%7B%286%5E2%29%5E1%7D-%5Csqrt%5B4%5D%7B%284%5E4%29%5E3%7D%5Cimplies%20%5Csqrt%5B3%5D%7B%287%5E2%29%5E3%7D%2B%5Csqrt%5B2%5D%7B%286%5E1%29%5E2%7D-%5Csqrt%5B4%5D%7B%284%5E3%29%5E4%7D%0A%5C%5C%5C%5C%5C%5C%0A7%5E2%2B6-4%5E3%5Cimplies%2049%2B6-64%5Cimplies%20-9)
to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.
Answer:
B. 1 and -2
Explanation:
First, add up the equations:
x-x+3y+y=4
4y=4
y=1
Now, plug y into any of the original equations above and solve for x:
-x+1=3
-x=2
x=-2
Therefore, the answer is B. 1 and -2
Answer:
$11/ day
Step-by-step explanation:
$44 /4 days
Divide the top and bottom by 4
44/4 = 11
$11/ day
Answer:
3x cubed- 3x squared- 11x -22
Assume that the amount needed from the 5% solution is x and that the amount needed from the 65% solution is y.
We are given that, the final solution should be 42 ml, this means that:
x + y = 42 ...........> equation I
This can also be written as:
x = 42-y .......> equation II
We are also given that the final concentration should be 45%, this means that:
5% x + 65% y = 45% (x+y)
0.05x + 0.65y = 0.45(x+y)
We have x+y = 42 from equation I, therefore:
0.05x + 0.65y = 0.45(42)
0.05x + 0.65y = 18.9 .........> equation III
Substitute with equation II in equation III as follows:
0.05x + 0.65y = 18.9
0.05(42-y) + 0.65y = 18.9
2.1 - 0.05y + 0.65y = 18.9
0.6y = 18.9 - 2.1
0.6y = 16.8
y = 28 ml
Substitute with y in equation II to get x as follows:
x = 42-y
x = 42 - 28
x = 14 ml
Based on the above calculations:
amount of 5% solution = x = 14 ml
amount of 65% solution = y = 28 ml
The correct choice is:
The teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution.
