Answer:
A. ![\frac{1}{6^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%5E%7B2%7D%20%7D)
Step-by-step explanation:
A. ![\frac{1}{6^{2} } = \frac{1}{36} = 0.02777](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%5E%7B2%7D%20%7D%20%20%3D%20%5Cfrac%7B1%7D%7B36%7D%20%20%3D%200.02777)
B. ![6^{-3} = 0.004629](https://tex.z-dn.net/?f=6%5E%7B-3%7D%20%3D%200.004629)
C. ![\frac{1}{216} = 0.004629](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B216%7D%20%3D%200.004629)
D. ![\frac{1}{6^{3} } = \frac{1}{216} = 0.004629](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%5E%7B3%7D%20%7D%20%20%3D%20%5Cfrac%7B1%7D%7B216%7D%20%3D%200.004629)
It should be noted that some of the examples of situations of transformation where the image of the object appear different but represent the same thing include:
-
The rigid transformation of geometric and three-dimensional shapes.
- The transformation of bank notes to an electronic form of money.
- The transformation of liquid water to ice and steam.
- The transformation of cube sugar to granulated sugar.
<h3>Analyzing transformations.</h3>
It should be noted that algebraic operations can be performed on polynomials in order to give equivalent expressions that have the same value.
In the example given above, it can be deduced that even though, there is a transformation as the things appear different, but its meaning remains the same.
Learn more about transformation on:
brainly.com/question/26068179
Answer:
LL
Step-by-step explanation:
You are given two congruent legs: BC≅BC and AC≅DC. The fact that it is a right triangle lets you invoke the LL theorem of congruence for right triangles. (This is a special case of the SAS theorem, where the A is 90°.)
Given:
![\begin{gathered} x^2-3x+2=0 \\ x^2+2x-15=0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E2-3x%2B2%3D0%20%5C%5C%20x%5E2%2B2x-15%3D0%20%5Cend%7Bgathered%7D)
Required:
We need to find the solution by Vièta's theorem.
Explanation:
Compare 1st equation with
![ax^2+bx+c=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3D0)
we get
![\begin{gathered} a=1 \\ b=-3 \\ c=2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%3D1%20%5C%5C%20b%3D-3%20%5C%5C%20c%3D2%20%5Cend%7Bgathered%7D)
Vièta's theorem is
![\begin{gathered} x_1+x_2=-\frac{b}{a} \\ x_1x_2=\frac{c}{a} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_1%2Bx_2%3D-%5Cfrac%7Bb%7D%7Ba%7D%20%5C%5C%20x_1x_2%3D%5Cfrac%7Bc%7D%7Ba%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} x_1+x_2=3 \\ x_1x_2=2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_1%2Bx_2%3D3%20%5C%5C%20x_1x_2%3D2%20%5Cend%7Bgathered%7D)
now solve this equation and we get
![\begin{gathered} x_1=1 \\ x_2=2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_1%3D1%20%5C%5C%20x_2%3D2%20%5Cend%7Bgathered%7D)
because addition of 1 and 2 is 3 and multiplication is 2
Now for 2nd equation
![\begin{gathered} a=1 \\ b=2 \\ c=-15 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20a%3D1%20%5C%5C%20b%3D2%20%5C%5C%20c%3D-15%20%5Cend%7Bgathered%7D)
apply Vièta's theorem
![\begin{gathered} x_1+x_2=-2 \\ x_1x_2=-15 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_1%2Bx_2%3D-2%20%5C%5C%20x_1x_2%3D-15%20%5Cend%7Bgathered%7D)
by this
![\begin{gathered} x_1=3 \\ x_2=-5 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x_1%3D3%20%5C%5C%20x_2%3D-5%20%5Cend%7Bgathered%7D)
because addition of 3 and -5 is -2 and multiplication is -15