1
:x
2
+y
2
−6x−9y+13=0
(x−3)
2
+(y−
2
9
)
2
−9−
4
81
+13=0
(x−3)
2
+(y−
2
9
)
2
=
4
65
Here,
r
1
=
2
65
C
1
=(3,
2
9
)
Equation of another circle-
S
2
:x
2
+y
2
−2x−16y=0
(x−1)
2
+(y−8)
2
−1−64=0
(x−1)
2
+(y−8)
2
=65
Here,
r
2
=
65
C
2
=(1,8)
Distance between the centre of two circles-
C
1
C
2
=
(3−1)
2
+(8−
2
9
)
2
C
1
C
2
=
4+
4
49
=
2
65
∣r
2
−r
1
∣=
∣
∣
∣
∣
∣
∣
65
−
2
65
∣
∣
∣
∣
∣
∣
=
2
65
∵C
1
C
2
=∣r
1
−r
2
∣
Thus the two circles touches each other internally.
Since the circle touches each other internally. The point of contact P divides C
1
C
2
externally in the ratio r
1
:r
2
, i.e.,
2
65
:
65
=1:2
Therefore, coordinates of P are-
⎝
⎜
⎜
⎜
⎜
⎜
⎛
1−2
1(1)−2(3)
,
1−2
1(8)−2(
2
9
)
⎠
⎟
⎟
⎟
⎟
⎟
⎞
=(5,1)
Therefore,
Equation of common tangent is-
S
1
−S
2
=0
(5x+y−6(
2
x+5
)−9(
2
y+1
)+13)−(5x+y−2(
2
x+5
)−16(
2
y+1
))=0
2
−6x−9y−13
+x+8y+13=0
4x−7y−13=0
Hence the point of contact is (5,1) and the equation of common tangent is 4x−7y−13=0.
9514 1404 393
Answer:
- annually: 9.01 years
- monthly: 8.69 years
- daily: 8.67 years
- continuously: 8.66 years
Step-by-step explanation:
For interest compounded in discrete intervals, the formula is ...
A = P(1 +r/n)^(nt)
We want to find t for P=1 and A=2, so we have ...
2 = (1 +r/n)^(nt)
ln(2) = nt·ln(1+r/n)
t = ln(2)/(n·ln(1+r/n))
A table of values for r=0.08 is attached.
__
For continuous compounding, the formula is ...
A = Pe^(rt)
t = ln(A/P)/r = ln(2)/0.08 ≈ 8.66434 . . . . years
__
- annually: 9.01 years
- monthly: 8.69 years
- daily: 8.67 years
- continuously: 8.66 years
Answer:
A
Step-by-step explanation:
Greetings! Hope this helps!
Answer
3 + 2h = 7
-3 -3
2h = 4
/2 /2
h = 2
Have a good day!
_______________
A brainliest would help tons! :D
You must mean the depth of the submarine.
to make sum = zero the depth of the sub must be -1600 ft..
