One of my factors is 9. The sum of my product of my two factors is 13. What inumber am I?
1 answer:
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Given:
The center of the circle = (-2,1).
Circle passes through the point (-5,3).
To find:
The equation of the circle.
Solution:
Radius is the distance between the center of the circle and any point on the circle. So, radius of the circle is the distance between the points (-2,1) and (-5,3).
On further simplification, we get
The standard form of a circle is:
Where, (h,k) is the center of the circle and r is the radius of the circle.
Substitute h=-2, k=1 and .
Therefore, the equation of the circle is .
Answer:
Elements are of the form
(i)
(ii)
(iii)
(iv)
(v) (a,b) where a>b=
(vi) [a,b] where a>b=
Step-by-step explanation:
Given intervals are,
(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.
To show all its elements,
(i) [a,a]
Imply the set including aa from left as well as right side.
Its elements are of the form.
Since there is a singleton element a of real numbers, this set is empty.
Because there is no increment so if then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].
(ii) [a,a)
This means given interval containing a by left and exclude a by right.
Its elements are of the form.
Since there is a singleton element a of real numbers withis the set, this set is empty.
Because there is no increment so if then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).
(iii) (a,a]
It means the interval not taking a by left and include a by right.
Its elements are of the form.
Since there is a singleton element a of real numbers, this set is empty.
Because there is no increment so if then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].
(iv) (a,a)
Means given set excluding a by left as well as right.
Since there is a singleton element a of real numbers, this set is empty.
Its elements are of the form.
Because there is no increment so if then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).
(v) (a,b) where a>b.
Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.
That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,
e.t.c
So this set is connected and we know singletons are connected in . Hence given set is empty.
(vi) [a,b] where .
Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.
That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,
e.t.c
So this set is connected and we know singletons are connected in . Hence given set is empty.