Answer:
A) reflection across the y-axis
Step-by-step explanation:
If 
, then this says that two 
-coordinates are equal for opposite values of 
.
Let 
a point on 
.
Then 
.
We also know that 
and therefore 
is also a point on the graph.
If you graph [/tex](-a,b)[/tex] and you will see they are symmetrical to each about the -axis.
Example if given both and 
, then 
. This means both (2,3) and (-2,3) are points on the graph.
Here is what those two points look like on a Cartesian Plane (please see graph in picture).
Answer:
520 feet
Step-by-step explanation:
The area=16900
Area formula=b^2
b^2=16900
square root of 16900=130
130*4=520
Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
![(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2](https://tex.z-dn.net/?f=%28x-%28a%2Bbi%29%29%2A%28x-%28a-bi%29%29%3D%5C%5C%28x-a-bi%29%2A%28x-a%2Bbi%29%3D%5C%5C%28%5Bx-a%5D-bi%29%2A%28%5Bx-a%5D%2Bbi%29%3D%5C%5C%28x-a%29%5E2-%28bi%29%5E2%3D%5C%5C%28x-a%29%5E2-b%5E2%28-1%29%3D%5C%5C%28x-a%29%5E2%2Bb%5E2)
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
Length × width=area
length+width + height
Answer: The proof is mentioned below.
Step-by-step explanation:
Here, Δ ABC is isosceles triangle.
Therefore, AB = BC
Prove: Δ ABO ≅ Δ ACO
In Δ ABO and Δ ACO,
∠ BAO ≅ ∠ CAO ( AO bisects ∠ BAC )
∠ AOB ≅ ∠ AOC ( AO is perpendicular to BC )
BO ≅ OC ( O is the mid point of BC)
Thus, By ASA postulate of congruence,
Δ ABO ≅ Δ ACO
Therefore, By CPCTC,
∠B ≅ ∠ C
Where ∠ B and ∠ C are the base angles of Δ ABC.
