Secant jkl and jmn are drawn to circle o from an external point ,j. if jk=8,lk=4 and jm=6 what is the length of jn answer
1 answer:
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 is
jn=16
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 is
jn=16
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Answer: The correct option is second, i.e. ,"2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".
Explanation:
The given equation is,
It can be written as,
Find the zeros of the equation. Equation the function equal to 0.
So, the three zeros are 0, 1 and 2.
The graph of the equation is shown below.
From the given graph it is noticed that the enclosed by the curve and x- axis is lies between 0 to 2, but the area from 0 to 1 lies above the x-axis and area from 1 to 2 lies below the x-axis. So the function will be negative from 1 to 2.
The area enclosed by curve and x-axis is,
From the graph it is noticed that the area from 0 to 1 is symmetric or same as area from 1 to 2. So the total area is the twice of area from 0 to 1.
Therefore, The correct option is "2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".
