4.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333repeating
See the picture attached to better understand the problem
we know that
If two secant segments are drawn to a <span>circle </span><span>from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
</span>so
jl*jk=jn*jm------> jn=jl*jk/jm
we have
<span>jk=8,lk=4 and jm=6
</span>jl=8+4----> 12
jn=jl*jk/jm-----> jn=12*8/6----> jn=16
the answer isjn=16
Answer:
?
Step-by-step explanation:
so first you have to do the formula then it would give you the answer
A.) Simplifying the equation would lead you to 15x^2-3x+9
b.) You know your answer is correct because you're adding the two polynomials together. 9x^2+6x^2 is 1tx^2. Now you have 2x-5x and since the negative is bigger, you get -3x. Then 5+4 is 9. You have no like terms therefore your answer is 15x^2-3x+9
Answer:
Therefore the rate change of height is
m/s.
Step-by-step explanation:
Given that a vertical cylinder is leaking water at rate of 4 m³/s.
It means the rate change of volume is 4 m³/s.

The radius of the cylinder remains constant with respect to time, but the height of the water label changes with respect to time.
The height of the cylinder be h(say).
The volume of a cylinder is 


Differentiating with respect to t.

Putting the value 



The rate change of height does not depend on the height.
Therefore the rate change of height is
m/s.