Answer:
ur mom gay
Step-by-step explanation:
Answer:
8 marbles
Step-by-step explanation:
Since Carl has 7 times as many marbles as Bill (who is the base unit), Carl's number of marbles is equal to 7 units.
7 units = 56
1 unit =
Bill has 8 marbles.
Answer:
I think it is C i could be wrong
Step-by-step explanation:
Point-slope form of a line: we need a point (x₀,y₀) and the slope "m".
y-y₀=m(x-x₀)
slope intercept form :
y=m+b
m=slope
If the line is parallel to y=2/3 x-0, the line will have the same slope, therefore the slope will be: 2/3.
Data:
(8,4)
m=2/3
y-y₀=m(x-x₀)
y-4=2/3(x-8)
y-4=2/3 x-16/3
y=2/3 x-16/3+4
y=2/3 x-4/3 (slope intercept form)
Answer: The equation of the line would be: y=2/3 x-4/3.
if we have the next slope "m",then the perendicular slope will be:
m´=-1/m
We have this equation: y=2/3 x+0; the slope is: m=2/3.
The perpendicular slope will be: m`=-1/(2/3)=-3/2
And the equation of the perpendicular line to : y=2/3 x+0, given the point (8,4) will be:
y-y₀=m(x-x₀)
y-4=-3/2 (x-8)
y-4=-3/2 x+12
y=-3/2x + 12+4
y=-3/2x+16
answer: the perpendicular line to y=2/3 x+0 , given the point (8,4) will be:
y=-3/2 x+16
This response is based upon your having had some background in calculus. "dx" is not introduced before that.
Take a look at the sample function y = f(x) = x^2 + 9. Here x is the independent variable; the dependent variable y changes with x.
Now, for a big jump: we consider finding the area under a curve (graph) between x = a and x = b. We subdivide that interval [a,b] into n vertical slices of area. Each of those slices has its own area: f(x)*dx, where dx represents the width of such subarea. f(x)*dx is the actual subarea. To find the total area under the curve f(x) between x= a and x = b, we add up all of these individual subareas between x = a and x = b. Note that the subinterval width is
b-a
dx = ---------- , and that dx becomes smaller and smaller as the number of
n subintervals increases.
Once again, this all makes sense only if you've begun calculus (particularly integral calculus). Do not try to relate it to earlier math courses.
