Solution :
Consider quadrilateral ABCD is a parallelogram. The parallelogram have diagonals AC and DB.
So in the given quadrilateral ABCD, let the diagonal AC and diagonal DB intersects at a point E.
Thus in the quadrilateral ABCD we see that :
1. AC and DB are the diagonals of quadrilateral ABCD.
2. Angle DCE is congruent to angle BAE and angle CDE is congruent to angle ABE. (they are alternate interior angles)
3. Line DC is congruent to line AB. (opposites sides are congruent in a parallelogram )
4. Angle ABE is congruent to angle CDE. (Angle side angle)
5. Line AE is congruent to line EC. And line DE is congruent to line EB. (CPCTC)
Thus we see that if the diagonals of a 
, then the quadrilateral is a parallelogram.
Answer:
(6a + 5b)(6a - 5b)
Step-by-step explanation:
Rewrite the expression in the form of a² - b²:
(6a)² - (5b)²
Use the difference of squares (a² - b² = (a + b)(a - b)):
(6a + 5b)(6a - 5b)
Multiply both sides by 2.3, ( tan 6° x 2.3 = f )
Regroup terms, 2.3 tan 6 = f.
F = 2.3 tan 6°
So the answer is tan 6 = tan 6 I think
Answer:
Step-by-step explanation:
First, you should put the axis of symetry as 5, -7 because f(x)= (x-h)^2+k and h is 5 and k is -7. the axis of symetry is x=5 because that is the "folding point" of the graph. Next, you would have to plug In 4 and 6 as your helping points to tell you where the other points of the function are going to be.
Plug 4 in and you get -6. plug 6 in and you get -6. plot the points (4, -6) and (6,-6).
Hope this helps!
:)
Answer:
B.
Step-by-step explanation:
the denominator you want is 16 so 4(4) is 16, which makes 3(4) 12 so 12/16. 7/16 is already perfect. 8(2) is 16 therefore 5(2) is 10 which makes 10/16.
