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:
1200
Explanation:
Order does not matter, if we said xyz order, it would still not make a difference if it was zyx or yzx hence we use the combination formula:
nCr = n! / r! * (n - r)!
where n= total number of items
r= number of items chosen at a time
Combinations are used when the order of events do not matter in calculating the outcome.
We calculate using the formula:
(30×20×12)÷3!=1200
There are therefore 1200 ways for the three distinct items to not be in same row or column
Answer:
40
Step-by-step explanation:
first do what is in the parentheses (4*4) which is 16 then rewrite the whole equation. 71-5(3)-16. then you'll want to multiply 5 times 3 because if there is no sign before the parentheses then it is automatically multiplication. 5 times 3 is 15. rewrite the equation again. 71-15-16 then simply subtract.
Hope this helps! :)
Answer:
19.2
Step-by-step explanation:
<u>1st Case:</u>
4 and 5 are legs of the right triangle.
Using the pythagorean therom: a^2+b^2=c^2
We can say that 4^2+5^2=x^2
16+25=x^2
41=x^2
x=√41
√41 is about 6.4
x=6.4
<u>2nd Case</u>
5 is the hypotenuse of the right triangle and 4 is the legs.
Using the pythagorean therom: a^2+b^2=c^2
We can say that 4^2+x^2=5^2
16+x^2=25
x^2=9
x=3
<u>Final Step</u>
We need to multiply the two possible lengths for x. So for case 1 the length of x was 6.4 and for case two the length was 3. 6.4*3=19.2
Anwser: <u>19.2</u>
The numbers are 4 and 2! 4 plus 2 equals 6! and 6 plus 4 is equal 10!
