Answer:
<h3>
Second option: ![b\sqrt{a}](https://tex.z-dn.net/?f=b%5Csqrt%7Ba%7D)
</h3>
Step-by-step explanation:
You need to remember the following property:
![\sqrt[n]{a^n} =a^{\frac{n}{n}}=a^1=a](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5En%7D%20%3Da%5E%7B%5Cfrac%7Bn%7D%7Bn%7D%7D%3Da%5E1%3Da)
You have the following expression provided:
![\sqrt{ab^2}](https://tex.z-dn.net/?f=%5Csqrt%7Bab%5E2%7D)
Applying the property mentioned before, you can rewrite it:
![=a^{\frac{1}{2}}b^{\frac{2}{2}}](https://tex.z-dn.net/?f=%3Da%5E%7B%5Cfrac%7B1%7D%7B2%7D%7Db%5E%7B%5Cfrac%7B2%7D%7B2%7D%7D)
Therefore, you get:
![=a^{\frac{1}{2}}b^1=b\sqrt{a}](https://tex.z-dn.net/?f=%3Da%5E%7B%5Cfrac%7B1%7D%7B2%7D%7Db%5E1%3Db%5Csqrt%7Ba%7D)
You can observe that this matches with the second option.
Yes, it is less then 1/2 of a tablespoon.
The value of x<em> </em>in the polynomial fraction 3/((x-4)•(x-7)) + 6/((x-7)•(x-13)) + 15/((x-13)•(x-28)) - 1/(x-28) = -1/20 is <em>x </em>= 24
<h3>How can the polynomial with fractions be simplified to find<em> </em><em>x</em>?</h3>
The given equation is presented as follows;
![\frac{3}{(x - 4) \cdot (x - 7) } + \frac{6}{(x - 7) \cdot (x - 13) } + +\frac{15}{(x - 13) \cdot (x - 28) } - \frac{1}{(x - 28) } = - \frac{1}{20}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B%28x%20-%204%29%20%5Ccdot%20%28x%20-%207%29%20%7D%20%20%2B%20%5Cfrac%7B6%7D%7B%28x%20-%207%29%20%5Ccdot%20%28x%20-%2013%29%20%20%20%7D%20%20%2B%20%2B%5Cfrac%7B15%7D%7B%28x%20-%2013%29%20%5Ccdot%20%28x%20-%2028%29%20%7D%20-%20%5Cfrac%7B1%7D%7B%28x%20-%2028%29%20%20%7D%20%3D%20%20-%20%20%5Cfrac%7B1%7D%7B20%7D%20)
Factoring the common denominator, we have;
![\frac{3\cdot(x - 13) \cdot(x - 28) + 6 \cdot(x - 4) \cdot(x - 28) + 15 \cdot(x - 4) \cdot(x - 7) - (x - 4) \cdot (x - 7)\cdot(x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)} + = - \frac{1}{20}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3%5Ccdot%28x%20-%2013%29%20%5Ccdot%28x%20-%2028%29%20%2B%206%20%5Ccdot%28x%20-%204%29%20%5Ccdot%28x%20-%2028%29%20%20%2B%2015%20%5Ccdot%28x%20-%204%29%20%5Ccdot%28x%20-%207%29%20%20-%20%28x%20-%204%29%20%5Ccdot%20%28x%20-%207%29%5Ccdot%28x%20-%2013%29%7D%7B%28x%20-%204%29%20%5Ccdot%20%28x%20-%207%29%5Ccdot%28x%20-%2013%29%20%5Ccdot%28x%20-%2028%29%7D%20%20%20%2B%20%3D%20%20-%20%20%5Cfrac%7B1%7D%7B20%7D%20)
Simplifying the numerator of the right hand side using a graphing calculator, we get;
By expanding and collecting, the terms of the numerator gives;
-(x³ - 48•x + 651•x - 2548)
Given that the terms of the numerator have several factors in common, we get;
-(x³ - 48•x + 651•x - 2548) = -(x-7)•(x-28)•(x-13)
Which gives;
![\frac{-(x - 7) \cdot(x - 28)\cdot (x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)} + = - \frac{1}{20}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-%28x%20-%207%29%20%5Ccdot%28x%20-%2028%29%5Ccdot%20%28x%20-%2013%29%7D%7B%28x%20-%204%29%20%5Ccdot%20%28x%20-%207%29%5Ccdot%28x%20-%2013%29%20%5Ccdot%28x%20-%2028%29%7D%20%20%20%2B%20%3D%20%20-%20%20%5Cfrac%7B1%7D%7B20%7D%20)
Which gives;
![\frac{-1}{(x - 4)} + = - \frac{1}{20}](https://tex.z-dn.net/?f=%20%5Cfrac%7B-1%7D%7B%28x%20-%204%29%7D%20%20%20%2B%20%3D%20%20-%20%20%5Cfrac%7B1%7D%7B20%7D%20)
x - 4 = 20
Therefore;
Learn more about polynomials with fractions here:
brainly.com/question/12262414
#SPJ1
-6+2 is -4. hope I helped!
Answer : Jeni's displacement from the starting point is, 111.8 meter.
Step-by-step explanation :
To calculate the Jeni's displacement from the starting point we are using Pythagoras theorem in ΔABC :
![(Hypotenuse)^2=(Perpendicular)^2+(Base)^2](https://tex.z-dn.net/?f=%28Hypotenuse%29%5E2%3D%28Perpendicular%29%5E2%2B%28Base%29%5E2)
![(AC)^2=(AB)^2+(BC)^2](https://tex.z-dn.net/?f=%28AC%29%5E2%3D%28AB%29%5E2%2B%28BC%29%5E2)
Given:
Side AB = 100 m
Side BC = 50 m
Now put all the values in the above expression, we get the value of side AC.
![(AC)^2=(AB)^2+(BC)^2](https://tex.z-dn.net/?f=%28AC%29%5E2%3D%28AB%29%5E2%2B%28BC%29%5E2)
![(AC)^2=(100)^2+(50)^2](https://tex.z-dn.net/?f=%28AC%29%5E2%3D%28100%29%5E2%2B%2850%29%5E2)
![AC=\sqrt{(100)^2+(50)^2}](https://tex.z-dn.net/?f=AC%3D%5Csqrt%7B%28100%29%5E2%2B%2850%29%5E2%7D)
![AC=111.8m](https://tex.z-dn.net/?f=AC%3D111.8m)
Thus, the Jeni's displacement from the starting point is, 111.8 meter.