From the diagram, we are given that triangle ABC and triangle ADC shares the same side AC, so it means both triangles have one congruent side.
We are also given that angle BAC and angle DAC are equal.
If we want to prove the triangles congruent using Angle-Angle-Side congruent postulate, we need to see one angle in each triangle that is congruent. We will need the information either: ACB = ACD or ABC = ADC
Correct answer: fourth option
Answer:
Option d) is correct
That is x equals plus or minus start fraction 11 over two end fraction
Step-by-step explanation:
Given quadratic equation is 
To write the given quadratic equation by using a difference-of-squares factoring method:
The above equation can be written as
The above equation is in the form of difference-of-squares
Therefore the given quadratic equation can be written in the form of difference-of-squares
by factoring method is
(which is in the form 
)
2x+11=0 or 2x-11=0
or 
or 
Therefore option d) is correct
That is x equals plus or minus start fraction 11 over two end fraction
Answer:
A just sub y of first equation into second then solve for X value
The answer is -2x^2-16x+8
Answer:
(x)^2 (y)^2
---------- + --------- = 1
4 3
Step-by-step explanation:
The standard equation for an ellipse is
(x-h)^2 (y-k)^2
---------- + --------- = 1
a^2 b^2
The center is at (h,k)
The vertices are at (h±a, k)
The foci are at (h±c,k )
Where c is sqrt(a^2 - b^2)
It is centered at the origin so h,k are zero
(x)^2 (y)^2
---------- + --------- = 1
a^2 b^2
The center is at (0,0)
The vertices are at (0±a, 0)
The foci are at (0±c,0 )
The vertices are (±2,0) so a =2
The foci is 1
c = sqrt(a^2 - b^2)
1 = sqrt(2^2 - b^2)
Square each side
1 = 4-b^2
Subtract 4 from each side
1-4 = -b^2
-3 = -b^2
3= b^2
Take the square root
b=sqrt(3)
(x)^2 (y)^2
---------- + --------- = 1
4 3
