![\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.
B - (5 x 8) - 3
A is adding 3, not subtracting.
C is basically 5 squared.
D is 5 lots of 11.
Answer:
top left:1/5, 2/7, 3/5, 5/7, 31/35
top right:2/5, 2/3, 7/10, 4/5, 9/10
bottom left:9/8, 15/8, 13/6, 9/4, 15/6
bottom right:13/10, 13/6, 11/5, 25/6, 21/5
Step-by-step explanation:
Hello!
Answer:
3/4x = -24
To isolate the "x" variable, divide both sides by 3/4 (multiply by its reciprocal, or 4/3)
3/4x · 4/3 = -24 · 4/3
x = (-24 · 4) / 3
x = -32.
Step-by-step explanation:
you are "hiding" some more information (like how much they made together).
without that we cannot calculate the actual values.
all I can do is set up the equations expressing the given relations between the parts of the total :
a = amount Alberto made
b = amount Benjamin made
c = amount Carlota made
b = 3×a
c = 2×b = 2× 3×a = 6×a
that's it.
your see ? now we need something that "ties" all 3 together, an equation of all 3 variables, where we can use the first 2 equations (by substitution) and then solve for the remaining third variable.
and that is missing.
if it is something like "together they made x", then we would have
a + b + c = x
a + 3a + 6a = x
10a = x
a = x/10
b and c we get then from the first 2 equations by simply using the calculated value of a :
b = 3×(x/10) = 3x/10
c = 6×(x/10) = 6x/10 = 3x/5
