The V shape comes from every possible "y" being positive, as an uninfluenced absolute value can never end up being negative.
Because it applies to every moment in our lives.
I see that but the problem is that it isn't right when I plug it into my class.
When x = 0 y = 6
y=0 x = 6
x =12 y = 6
x=-6 y=12
graph and connect those points....
It will be a ' V' shaped graph
How do I solve the equation then
Use the x and y axis
How do you THINK it is SUPPOSED to work??
You have everything you need to plug into the equation and solve for 'r' ..... {nl} 2.28x 10^19 = 6.67 x 10^(-11) x 8.68 x 10^25 x 6.59 x 10^19 / r^2
r^2 = 167.3385 x 10^14
r = 12.93593 x 10^7 m
r = 1.293593 x 10^8 m
r = 1.293593 x 10^5 KILOmeters
doesnt work how its suposed to
The \({x}^{y}\) button is an exponent button
9x-(2x+4)
\(9x-(2x+4)= 9x-2x -4\large=7x-4\) !
y-5=1/3(x-(-4)
\(y-5=\frac{1}{3(x-(-4))}\) is that right?
\(y-5=\frac{1}{3(x+4)}\)
\(y-5=\frac{1}{3x+12}\)
\(y=5+\frac{1}{3x+12}\) !
3.33.
-18 = -3 +v/3
-18+3 = v/3
-15x3 = v
v = -45
Depends on your calculator.
cos-1 is for when you dont know what the angle is but you are given the adjacent and hypotenuse values.
For example if the hypotenuse is 10 and the adjacent is 5 you would input.
cos-1(5/10) = 60 degrees
3-4cos(x)=7-(2-cos(x)
3-4cos(x)=7-2+cos(x)
3-4cos(x)=5+cos(x) |-cos(x)-3
-5cos(x)=2 | : (-5)
cos(x)=-2/5
x=113,58°
1+1/2+1/4+1/8+1/16+1/32+... ? → 2 (if the dots mean carry on to infinity)
.
Are you trying to compute this is MATLAB? and you would have to supply the value of b.
usually diff(a,b) implies a-b.
180 x 2
90x2x2
45x2x2x2
5x9x2x2x2
5x3x3x2x2x2
5x32x23
!
