Answer:
ABCD is a parallelogram.
Step-by-step explanation:
A parallelogram is a quadrilateral that has two parallel and equal pairs of opposite sides.
From the given diagram,
Given: AD = BC and AD || BC, then:
i. AB = DC (both pairs of opposite sides of a parallelogram are congruent)
ii. <ADC = < BCD and < DAB = < CBA
thus, AD || BC and AB || DC (both pairs of opposite sides of a parallelogram are parallel)
iii. < BAC = < DCA (alternate angle property)
iv. Join BD, line AC and BC are the diagonals of the quadrilateral which bisect each other. The two diagonals are at a right angle to each other.
v. <ADC + < BCD + < DAB + < CBA = 
(sum of angles in a quadrilateral equals 4 right angles)
Therefore, ABCD is a parallelogram.
Answer:
Step-by-step explanation:
We can add the two equations to eliminate the variable 
and solve for 
:
Now, we can use substitution to solve for :
(given)
Hope this helps :)
Your answer is 25. You do 40 times 5 divided by 8 or 8 times 5 is 40 and 5 times 5 is 25. Hope this helps
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
