Answer:
b. When you divide both sides by 2x = 6x it could lead us to think that there is no solution while, in fact, the solution is x = 0.
Step-by-step explanation:
The solution is correct up to the step 2x = 6x
2x = 6x
Subtract 2x from both sides.
0 = 4x
Divide both sides by 4.
x = 0
You cannot divide both sides by x since x could be zero, and in fact, it is.
Answer: b. When you divide both sides by 2x = 6x it could lead us to think that there is no solution while, in fact, the solution is x = 0.
2 is the base and (
![\sqrt[4][a^5 b^3]](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%5Ba%5E5%20b%5E3%5D)
) is the exponent.
the 4 is called an index and it means to fi d the fourth root of the expression under the (
![\sqrt[][text]) symbol.\\a fourth root is a factor that was multiplied four times. ex: 2*2*2*2=16. the fourth root of 16 is two. \\when you do square roots, you take the number representing the factor out of the radical. anything else that is not a square root stays under. ex: \\([tex]\sqrt[9][text])\\([tex]\sqrt[3*3*3][text]) there is one set of two 3's so\\([tex]3\sqrt[3][text])\\with a fourth root, you look for groups of four. the symbol of that group is placed outside the radical. anything else stays in.\\ ([tex]\sqrt[4][a^5 b^3]](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%5Btext%5D%29%20symbol.%5C%5Ca%20fourth%20root%20is%20a%20factor%20that%20was%20multiplied%20four%20times.%20ex%3A%202%2A2%2A2%2A2%3D16.%20the%20fourth%20root%20of%2016%20is%20two.%20%5C%5Cwhen%20you%20do%20square%20roots%2C%20you%20take%20the%20number%20representing%20the%20factor%20out%20of%20the%20radical.%20anything%20else%20that%20is%20not%20a%20square%20root%20stays%20under.%20ex%3A%20%5C%5C%28%5Btex%5D%5Csqrt%5B9%5D%5Btext%5D%29%5C%5C%28%5Btex%5D%5Csqrt%5B3%2A3%2A3%5D%5Btext%5D%29%20there%20is%20one%20set%20of%20two%203%27s%20so%5C%5C%28%5Btex%5D3%5Csqrt%5B3%5D%5Btext%5D%29%5C%5Cwith%20a%20fourth%20root%2C%20you%20look%20for%20groups%20of%20four.%20the%20symbol%20of%20that%20group%20is%20placed%20outside%20the%20radical.%20anything%20else%20stays%20in.%5C%5C%20%28%5Btex%5D%5Csqrt%5B4%5D%5Ba%5E5%20b%5E3%5D)
)
(
![\sqrt[4][a*a*a*a*a*b*b*b]](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%5Ba%2Aa%2Aa%2Aa%2Aa%2Ab%2Ab%2Ab%5D)
) there is one group of 4 a's so
(
![a\sqrt[4][a* b^3]](https://tex.z-dn.net/?f=a%5Csqrt%5B4%5D%5Ba%2A%20b%5E3%5D)
)
there really isn't anything else to do to simplify the expression
Compare the numbers using <, >, or =. 0.78 ___ 0.708 < > =
The two number are 39 and 13
<em><u>Solution:</u></em>
Let the two numbers be "a" and "b"
Let the larger number be "a" and the smaller number be "b"
<em><u>Given that, sum of two numbers is 52</u></em>
a + b = 52 ---------- eqn 1
<em><u>One number is 3 times as large as the other number</u></em>
Larger number = 3 times smaller number
a = 3b -------- eqn 2
<em><u>Let us solve eqn 1and eqn 2</u></em>
<em><u>Substitute eqn 2 in eqn 1</u></em>
3b + b = 52
4b = 52
b = 13
<em><u>Substitute b = 13 in eqn 2</u></em>
a = 3(13)
a = 39
Thus the two number are 39 and 13
