<span> x = 2<span> x= 0.0000 - 1.4142 i
</span><span> x= 0.0000 + 1.4142 i
</span></span>
Let's solve for x.<span><span><span><span>5k</span>x</span>+6</span>=<span><span>7k</span>x</span></span>
Step 1: Add -7kx to both sides.<span><span><span><span><span>5k</span>x</span>+6</span>+<span>−<span><span>7k</span>x</span></span></span>=<span><span><span>7k</span>x</span>+<span>−<span><span>7k</span>x</span></span></span></span><span><span><span>−<span><span>2k</span>x</span></span>+6</span>=0</span>
Step 2: Add -6 to both sides.<span><span><span><span>−<span><span>2k</span>x</span></span>+6</span>+<span>−6</span></span>=<span>0+<span>−6</span></span></span><span><span>−<span><span>2k</span>x</span></span>=<span>−6</span></span>
Step 3: Divide both sides by -2k.<span><span><span><span><span>−<span><span>2k</span>x</span></span><span>−<span>2k</span></span></span></span></span>=<span><span><span><span>−6</span><span>−<span>2k</span></span></span></span></span></span><span>x=<span><span>3/<span>k</span></span></span></span>
The gradient is the slope right?
If yes, then your answer is -4.
m=(y2-y1)/(x2-x1)
m=(10-2)/(-2-0)
m=-4
I would think that all but one point would be on the line. One way to approach this problem is to find the equation of the line based upon any two points chosen at random, and then determine whether or not the other points satisfy this equation. Next time, would you please enclose the coordinates of each point inside parentheses: (2.5,14), (2.25,12), and so on, to avoid confusion.
14-12
slope of line thru 1st 2 points is m = ---------------- = 2/0.25 = 8
2.50-2.25
What is the eqn of the line: y = mx + b becomes
14 = (8)(2.5) + b; find b:
14-20 = b = -6. Then, y = 8x - 6.
Now determine whether (12,1.25) lies on this line.
Is 1.25 = 8(12) - 6? Is 1.25 = 90? No. So, unless I've made arithmetic mistakes, (1.25, 5) does not lie on the line thru (2.5,14) and (2.25,12).
Why not work this problem out yourself using my approach as a guide?
14(3+1) - (3 - 7)
"double" - times by 2
"and" - addition
when using " first second and third" you just have to assign the numbers by the way it comes in the problem...
first number - 7
second number - 3
third number - 1
