Answer:
Step-by-step explanation:
First you have to find the medians which is when you put the numbers in number order and find the one in the middle.
Class A: 60,65,70,70,75,80,80,85,90,90
=77.5
Class B: 75,85,85,85,90,90,90,90,95,100
=90
That the class B is more advanced, and they probably studied.
Answer:151500
Step-by-step explanation:the is the answer
You plug -2 in for the function k(p) and add it to the function g(w), getting
(-2+3)*(-2-7)+(-2-5)^2=1*-9+49=40 for a - I challenge you to do B on your own!
Answer:
x value of vertical asymptote and y value of horizontal asymptote
Step-by-step explanation:
The graph of 1/x approaches infinity as x approaches 0 (the vertical asymptote)
As x gets either bigger or smaller, 1/x approaches the x-axis (from above on the positive side, from below on the negative side) (the horizontal asymptote)
Consider 1/(x-5) + 2, at what value of x does the graph 'go nuts' ?
When the bottom of the fraction becomes 0, x - 5 becomes 0 when x = 5, so the vertical asymptote of g(x) is at x=5
What value of y does f(x) approach as x gets more positive or more negative - as x gets bigger (as an example), y approaches 0
What y value does g(x) approach as x gets bigger? Well, as x gets big, 1/(x-5) gets small, approaching 0. The smallest 0 + 2 can get is 2, so y=2 is the horizontal asymptote
The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.
The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
</span>
Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C
