Question 5 of 8 In arithmetic, variables look like Answer here SUBMIT
1 answer:
Answer:
In arithmetic, variables look like letters.
Step-by-step explanation:
As we know that Mathematical arithmetic deals with numbers and certain properties with them.
Some of the main properties between numbers are as follows:
- Addition
- Subtraction
- Multiplication
- Division
Variables do not remain constant, so we basically assign a number to a variable with an equal operator denoted by '='.
For example, let us assume a variable x being equal to 7. so it can written as:
Let suppose the value of the variable x=7 changes to x=10, then the new value will be:
Therefore, the correct answer is:
In arithmetic, variables look like letters.
You might be interested in
Unfortunately the portion of the graph shown does not show the locations of points A and B, nor does it show sufficient parts of the curve.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.
Fortunately, to do part C, it is not necessary to use the graph of the function, nor the locations of points A & B, since we know that t=-1 at A.
The solution is as follows:
We know that the tangent line has the equation
l : 2x -5y -9 = 0 ...................(1)
and that the parametric equation is
x = t^3 -8t; y = t^2 ..............(2)
We can obtain the intersections of the tangent line (1) with the original curve (2) by substituting (2) in (1), giving
f(t) = 2*(t^3-8*t) -5*(t^2)-9 = 0
or on simplification,
f(t) = 2t^3 -5t^2 -16t -9 = 0 ....................(3) parametric equation of intersection points
Solution of which gives t=9/2 or t=-1.
Since t=-1 is the given tangent point, we conclude that t=9/2 is point B.
Translating into rectangular coordinates for t=9/2, we get
x=t^3-8t=(9/2)^3-8(9/2)= 441/8 = 55.125
y=t^2=(9/2)^2= 81/4 = 20.25
Therefore the coordinates of B:
t=9/2, or (55.125, 20.25)
Attached is the graphics of f(t), showing that t=-1 is a tangent line, and that the solutions are t=-1, or t=9/2.