ANSWER
Yes it is very true
<u>EXPLANATION</u>
If the two equations intersect at 
then this point must satisfy the two equations.
We substitute in to erquation (1)
We now substitute in to erquation (2) also
Since the point satisfy all the two equations, it is true that they intersect at 
Step-by-step explanation:
h = height of rectangle (should be "length", but ok)
w = width of rectangle
w = h + 97
the diagonal of the rectangle is the Hypotenuse (baseline) of a right-angled triangle with the legs being one height and one width side of the rectangle.
so, we use Pythagoras
c² = a² + b²
with c being the Hypotenuse (the side opposite of the 90° angle).
diagonal² = h² + (h + 97)² = h² + h² + 2×97h + 97²
113² = 12769 = 2h² + 194h + 9409
3360 = 2h² + 194h
1680 = h² + 97h = (h + 97/2)² - 9409/4
(completing the square)
6720/4 + 9409/4 = (h + 97/2)²
16129/4 = (h + 97/2)²
127/2 = h + 97/2
30/2 = h = 15
the height is 15.0 in
estimate value of :- 2.99548 is 3
estimate value of :- 1.8342 is 2
So 3•2=6
so 6 is your answer
Consider this option:
1. rule: Circle is the locus of the points which are on the same distance from the given point. Common view of the equation is: (x-a)²+(y-b)²=r², where r - the distance from the given point to the described locus, a&b - coordinates of the given point.
2. using item 1:
(x+2)²+(y-1)²=3² or (x+2)²+(y-1)²=9.
Divides el (1/2) q es 2 pq 2*1=2 y después multiplicas 2*2=4 entonces la respuesta es 4 y es base creo
