Hello from MrBillDoesMath!
Answer: infinite solutions
Discussion: I did a double take on this one but the left hand is
a + 3 + 2a = 3a +3
and the right hand side is
-1 + 3a + 4 = 3a + (4-1) = 3a + 3
The left and right sides of the equation are identical for all "a", i.e. for infinitely many "a" values.
Regards, MrB.
Answer:
A) None
Step-by-step explanation:
1) 
shoudnt neccesarily be a factor of nst, for example, if s = 3, t = 4, and n = 12, then both s and t are factors of n, but 
is not a factor of nst = 144.
2) 
shoudnt neccesarily be a factor of nst. Let s be 4, let t be 6, and let n be 12. Then n is a factor of both s and t, but 
is not a factor of nst = 12*24. In fact, it is a greater number.
3) Again, s+t isnt necessarily a factor of nst, let s be 2 and t be 3. Then both s and t are factor of n = 12. However 5 = s+t is not a factor of nst = 72.
So, neither of the three options is guaranteed to be a factor of nst. In fact, for s = 4, t = 6, and n = 12, none of the three options are valid.
Answer:
3
Step-by-step explanation:
j(2) means that x equals 2...
the first inequality in the function says that x is greater than or equal to 2
the second inequality states that x is greater than 2(not equal to) and is less than or equal to 5
the third one states that x is greater than 5
x applies to the first inequality since we know that x can be equal to 2
Now, to the number before the if
there is a 3 before the if in the first inequality
This means that if x is greater than or equal to 2 the y or the function would equal 3
therefore, j(2) would equal 3
I hope this helps!!!
In this case we use intersecting secant theorem
According to this theorem, GF*GE = GH*GA
So,we plug in the values
(x-3)*(x-3+9) =(x-2)*(x-2+6)
(x-3)*(x+6) =(x-2)*(x+4)
x^2 +6x -3x -18 =x^2 +4x -2x -8
x^2 +3x -18 =x^2 +2x -8
x^2 +3x -18 -x^2 -2x +8 =0
x -10 =0
x=10
So,FG =x-3 = 10-3 =7
