Given:
The x and y axis are tangent to a circle with radius 3 units.
To find:
The standard form of the circle.
Solution:
It is given that the radius of the circle is 3 units and x and y axis are tangent to the circle.
We know that the radius of the circle are perpendicular to the tangent at the point of tangency.
It means center of the circle is 3 units from the y-axis and 3 units from the x-axis. So, the center of the circle is (3,3).
The standard form of a circle is:
Where, (h,k) is the center of the circle and r is the radius of the circle.
Putting 
, we get
Therefore, the standard form of the given circle is .
Answer:
6 ways - ab,ac, ba,bc and ca,cb
Step-by-step explanation:
We are given 3 letters a,b,c.
We can choose the first letter in 1 out of 3 ways.
Once the first letter has been chosen without replacement, we have two letters remaining. Another letter can be chosen from the 2 remaining letters in 2 ways. So the total number of ways of choosing the two letters is 3*2 = 6.
Listing out the possible set of choices:
Options include: ab,ac, ba,bc and ca, cb
20 and 10, 20 is 2 times 10 and if you add them together its 30
More than means addition.
C is the correct option.
= (2x + 4) + 9 - 4y + 3x
= 2x + 4 + 9 - 4y + 3x
= 5x + 13 - 4y
= 5x - 4y + 13
Let us first define Hypotenuse Leg (HL) congruence theorem:
<em>If the hypotenuse and one leg of a right angle are congruent to the hypotenuse and one leg of the another triangle, then the triangles are congruent.</em>
Given ACB and DFE are right triangles.
To prove ΔACB ≅ ΔDFE:
In ΔACB and ΔDFE,
AC ≅ DF (one side)
∠ACB ≅ ∠DFE (right angles)
AB ≅ DE (hypotenuse)
∴ ΔACB ≅ ΔDFE by HL theorem.
