<u>Answer:</u>
1/4
<u>Step-by-step explanation:</u>
We are to find the scale factor of the dilation that maps the pre-image of triangle ABC with vertices A(2, 5), B(6, 10) and C(9, −1) to the image triangle A'B'C' with vertices A' (0.5, 1.25), B' (1.5, 2.5), C' (2.25, −0.25).
Center of dilation is at the origin.
To find the scale factor, we will divide the corresponding vertices of the image and pre-image.
A(2, 5) ---> A' (0.5, 1.25) = ![\frac{0.5}{2} , \frac{1.25}{5}=(\frac{1}{4} , \frac{1}{4})](https://tex.z-dn.net/?f=%5Cfrac%7B0.5%7D%7B2%7D%20%2C%20%5Cfrac%7B1.25%7D%7B5%7D%3D%28%5Cfrac%7B1%7D%7B4%7D%20%2C%20%5Cfrac%7B1%7D%7B4%7D%29)
B(6, 10) ---> B' (1.5, 2.5) = ![\frac{1.5}{6} , \frac{2.5}{10}=(\frac{1}{4} , \frac{1}{4})](https://tex.z-dn.net/?f=%5Cfrac%7B1.5%7D%7B6%7D%20%2C%20%5Cfrac%7B2.5%7D%7B10%7D%3D%28%5Cfrac%7B1%7D%7B4%7D%20%2C%20%5Cfrac%7B1%7D%7B4%7D%29)
C(9, −1) ---> C' (2.25, −0.25) = ![\frac{2.25}{9} , \frac{-0.25}{-1}=(\frac{1}{4} , \frac{1}{4})](https://tex.z-dn.net/?f=%5Cfrac%7B2.25%7D%7B9%7D%20%2C%20%5Cfrac%7B-0.25%7D%7B-1%7D%3D%28%5Cfrac%7B1%7D%7B4%7D%20%2C%20%5Cfrac%7B1%7D%7B4%7D%29)
Therefore, the scale factor of the dilation is 1/4.
What’s the question??????????
Answer:
Step-by-step explanation:
Hi, there you have to use the exponents rule
![(a^{b})^{c}= a^{bc}](https://tex.z-dn.net/?f=%28a%5E%7Bb%7D%29%5E%7Bc%7D%3D%20a%5E%7Bbc%7D)
![6^{2*4}](https://tex.z-dn.net/?f=6%5E%7B2%2A4%7D)
=
1679616
Hope this helps :)
Answer:
Only d) is false.
Step-by-step explanation:
Let
be the characteristic polynomial of B.
a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)
b) Remember that
. 0 is a root of p, so we have that
.
c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.
d) det(B)=0 by part c) so B is not invertible.
e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.