Answer:
The two triangles are related by Side-Side-Side (SSS), so the triangles can be proven congruent.
Step-by-step explanation:
There are no angles that can be shown to be congruent to one another, so this eliminates all answer choices with angles (SSA, SAS, ASA, AAA, AAS).
This leaves you with either the HL (Hypotenuse-Leg) Theorem or SSS (Side-Side-Side) Theorem. We could claim that the triangles can be proven congruent by HL, however, we aren't exactly sure as to whether or not the triangles have a right angle. There is no indicator, and in this case, we cannot assume so.
This leaves you with the SSS Theorem.
First get the factors of 24: 2*2*2*3Then the factors of 9: 3*3 Comparing the factors, the GCF is 3 Then you can rewrite the expression: 3*8 + 3*3 At this point, I'm not sure whether what you mean is really distributive property or not since this case is more of a factoring. 3*(8+3)
Since f(x) is a polynomial with 3rd degree, then it will have 3 roots (zeroes)
One of them is real and the other two are complex conjugate roots
Since the real root is 4, then
x = 4
Since the complex root is (1 - i), then
The other root will be the conjugate of it (1 + i)
x = (1 - i)
x = (1 + i)
To find f(x) we will multiply the three factors of it
We can get the factors from the zeroes
![x=4](https://tex.z-dn.net/?f=x%3D4)
Subtract 4 from both sides
![\begin{gathered} x-4=4-4 \\ x-4=0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x-4%3D4-4%20%5C%5C%20x-4%3D0%20%5Cend%7Bgathered%7D)
The first factor is (x - 4)
![\begin{gathered} x=1-i \\ x-(1-i)=(1-i)-(1-i) \\ x-1+i=0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D1-i%20%5C%5C%20x-%281-i%29%3D%281-i%29-%281-i%29%20%5C%5C%20x-1%2Bi%3D0%20%5Cend%7Bgathered%7D)
The second factor is (x - 1 + i)
The third factor is (x - 1 - i)
![f(x)=(x-4)(x-1+i)(x-1-i)](https://tex.z-dn.net/?f=f%28x%29%3D%28x-4%29%28x-1%2Bi%29%28x-1-i%29)
We will multiply them to find f(x)
![\begin{gathered} (x-1+i)(x-1-i)= \\ x^2-x-ix-x+1+i+ix-i-(i^2)= \\ x^2-2x+1-(-1)= \\ x^2-2x+1+1= \\ x^2-2x+2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%28x-1%2Bi%29%28x-1-i%29%3D%20%5C%5C%20x%5E2-x-ix-x%2B1%2Bi%2Bix-i-%28i%5E2%29%3D%20%5C%5C%20x%5E2-2x%2B1-%28-1%29%3D%20%5C%5C%20x%5E2-2x%2B1%2B1%3D%20%5C%5C%20x%5E2-2x%2B2%20%5Cend%7Bgathered%7D)
Multiply it by (x - 4)
![\begin{gathered} f(x)=(x-4)(x^2-2x+2) \\ f(x)=x^3-2x^2+2x-4x^2+8x-8 \\ f(x)=x^3-6x^2+10x-8 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20f%28x%29%3D%28x-4%29%28x%5E2-2x%2B2%29%20%5C%5C%20f%28x%29%3Dx%5E3-2x%5E2%2B2x-4x%5E2%2B8x-8%20%5C%5C%20f%28x%29%3Dx%5E3-6x%5E2%2B10x-8%20%5Cend%7Bgathered%7D)
The answer is
The correct answer is d.We have the following system of linear equations:
(I)
![4x+7y=-14](https://tex.z-dn.net/?f=4x%2B7y%3D-14)
(II)
![8x+5y=8](https://tex.z-dn.net/?f=8x%2B5y%3D8)
Let's use the elimination method, then let's multiply the equation (1)
![\times(-2)](https://tex.z-dn.net/?f=%5Ctimes%28-2%29)
and subtracting (I) and (II):
(I)
![-2(4x+7y)=-2(-14)](https://tex.z-dn.net/?f=-2%284x%2B7y%29%3D-2%28-14%29)
∴
![-8x-14y=28](https://tex.z-dn.net/?f=-8x-14y%3D28)
(I)
![-8x-14y=28](https://tex.z-dn.net/?f=-8x-14y%3D28)
(II)
![8x+5y=8](https://tex.z-dn.net/?f=8x%2B5y%3D8)
____________________
(III)
![-9y=36](https://tex.z-dn.net/?f=-9y%3D36)
∴
![y=-4](https://tex.z-dn.net/?f=y%3D-4)
We can find the value of x by substituting y either in (I) or (II). Thus, from (I):
![4x+7(-4)=-14](https://tex.z-dn.net/?f=4x%2B7%28-4%29%3D-14)
∴
![4x-28=-14](https://tex.z-dn.net/?f=4x-28%3D-14)
∴
![4x=14](https://tex.z-dn.net/?f=4x%3D14)
∴
![x=\frac{14}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B14%7D%7B4%7D)
∴
![\boxed{x=\frac{7}{2}}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D%5Cfrac%7B7%7D%7B2%7D%7D)
Let's substitute the values of x and y into (I) and (2)
(I)
![4(\frac{7}{2})+7(-4)=14-28=-14](https://tex.z-dn.net/?f=4%28%5Cfrac%7B7%7D%7B2%7D%29%2B7%28-4%29%3D14-28%3D-14)
(II)
![8(\frac{7}{2})+5(-4)=28-20=8](https://tex.z-dn.net/?f=8%28%5Cfrac%7B7%7D%7B2%7D%29%2B5%28-4%29%3D28-20%3D8)
Finally the answer is: