Answer:
The length of segment QM' = 6
Step-by-step explanation:
Given:
Q is the center of dilation
Pre-image (original image) = segment LM
New image = segment L'M'
The length of LQ = 4
The length of QM = 3
The length of LL' = 4
The original image was dilated with scale factor = 2
QM' = ?
To determine segment QM', first we would draw the diagram obtained from the given information.
Find attached the diagram
When a figure is dilated, we would have similar shape in thus cars similar triangles.
Segment L'M' = scale factor × length of LM
Let LM = x
L'M' = 2x
Using similar triangles theorem, ratio of their corresponding sides are equal.
QM/LM = QM'/L'M'
3/x = QM'/2x
6x = QM' × x
Q'M' = 6
The length of segment QM' = 6
Answer:
I think this ans may help you
Answer:
Step-by-step explanation:
A. irrational
b. irrational
c. rational
d. irrational
e. rational
hope it helps
Answer:
![\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Csf%20%5Cbf%20%5C%20%5C%20f%28x%29%3D%28x-4i%29%28x%2B4i%29%28x%2B3%29%28x-5%29%20%5C%20%5C%20%7D)
Step-by-step explanation:
Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).
![\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}](https://tex.z-dn.net/?f=%5Ctext%7BMeaning%20that%20%7D%20%28x-4i%29%28x%2B4i%29%20%3Dx%5E2-%284i%29%5E2%3Dx%5E2%2B16%20%5Ctext%7B%20is%20a%20factor%20of%20f%28x%29.%7D)
The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.
![f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E4-2x%5E3%2Bx%5E2-32x-240%5C%5C%5C%5C%20%3D%28x%5E2%2B16%29%28x%5E2%2Bax-15%29%5C%5C%5C%5C%20%3Dx%5E4%2Bax%5E3-15x%5E2%2B16x%5E2%2B16ax-240)
We identify the coefficients for the like terms, it comes
a = -2 and 16a = -32 (which is equivalent). So, we can write in
.
![\\f(x)=(x^2+16)(x^2-2x-15)](https://tex.z-dn.net/?f=%5C%5Cf%28x%29%3D%28x%5E2%2B16%29%28x%5E2-2x-15%29)
The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.
![f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}](https://tex.z-dn.net/?f=f%28x%29%3D%28x%5E2%2B16%29%28x%5E2-2x-15%29%5C%5C%5C%5C%3D%28x%5E2%2B16%29%28x%5E2%2B3x-5x-15%29%5C%5C%5C%5C%3D%28x%5E2%2B16%29%28x%28x%2B3%29-5%28x%2B3%29%29%5C%5C%5C%5C%3D%5Cboxed%7B%28x%5E2%2B16%29%28x%2B3%29%28x-5%29%7D)
And we can write in ![\mathbb{C}](https://tex.z-dn.net/?f=%5Cmathbb%7BC%7D)
![f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cboxed%7B%28x-4i%29%28x%2B4i%29%28x%2B3%29%28x-5%29%7D)
Hope this helps.
Do not hesitate if you need further explanation.
Thank you