<span>two divided by the difference of a number and 3 minus 4 divided by a number plus 3, equals 8 times the reciprocal of the difference of the number squared and 9.
let x=number
2/(x-3)-4/(x+3)=8/(x^2-9)=8/(x+3)(x-3)
LCD: (x+3)(x-3)
2(x+3)-4(x-3)=8
2x+6-4x+12=8
-2x+18=8
-2x=-10
x=5</span>
Answer:
121
Step-by-step explanation:
The missing number is 121. I got this by doing 764-130-513=121. You would do this because you are trying to find the number that would allow 130 and 513 to add up to 764.
Another way to look at this problem is 513+130= 643
764-643=121
In conclusion, 513+121+130=764
Answer:
b. No porque su cociente es 0,5.
Answer:
(a) 4i times
(b) "i × n" times
Step-by-step explanation:
(a) Given the algorithm segment;
for i := 1 to 4, (Outer loop)
for j := 1 to i (Inner loop)
next j,
next i
The inner loop runs for i times, while the outer loop runs for 4 times.
The total times the inner loop would run when the entire algorithm is run is:
= i × 4
= 4i times
(b) Given the algorithm segment;
for i := 1 to n, (Outer loop)
for j := 1 to i (Inner loop)
next j,
next i
Where n is a set of positive integers.
The inner loop runs for "i" times, while the outer loop runs for "n" times.
The total times the inner loop would run when the entire algorithm is run is:
= i × n
= "i × n" times
The solution to this equation is C (6)
3x - 3 = 15
3x = 15 + 3
3x = 18
3/3 x = 18/3
x = 6
