$9.00 per hour. take 45 and divide by 5 you get 9.
Answer:
C.
Step-by-step explanation:
You are given 3x-4y<16 and we want to see which of the ordered pairs is a solution.
These ordered pairs are assumed to be in the form (x,y).
A. (0,-4)
?
3x-4y<16 with (x=0,y=-4)
3(0)-4(-4)<16
0+16<16
16<16 is not true so (0,-4) is not a solution of the given inequality.
B. (4,-1)?
3x-4y<16 with (x=4,y=-1)
3(4)-4(-1)<16
12+4<16
16<16 is not true so (4,-1) is not a solution of the given inequality.
C. (-3,-3)?
3x-4y<16 with (x=-3,y=-3)
3(-3)-4(-3)<16
-9+12<16
3<16 is true so (-3,-3) is a solution to the given inequality.
D. (2,-3)?
3x-4y<16 with (x=2,y=-3)
3(2)-4(-3)<16
6+12<16
18<16 is false so (2,-3) is not a solution to the given inequality.
Is this calculus or something?
Use the formula or complete the square.
The zeroes of the quadratic can be real and rational; real and irrational; complex conjugates.
If the quadratic is ax²+bx+c, x=(-b+√b²-4ac)/2a.
If b² > 4ac the solutions are real. If b²-4ac is a perfect square, the solutions are real and rational; otherwise they’re real but irrational.
If b² < 4ac the solutions are complex.
Answer:
Not really sure but 14.25 students seem to be the correct answer.
Step-by-step explanation:
114:8 = 14.25
