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Good to me.
B
Very good. Like Richie Cunningham is famous for saying on Happy Days:
I'M COOKING WITH GAS.
If you set x = 0 in the y = mx + c, you'll get the y intercept and y = 0 gives the x one.
Symmetry? try replacing x by -x. If the equation simplfies to the same again then the y axis is a line of symmetry.
eg. y = x^2 replace x with -x and we have y = (-x)^2 = x^2
If the replacement gives -y then the graph is rotationally symmetric aroung the origin.
eg y = x^3 replace x with -x and we have y = (-x)^3 = -x^3 = -y
You can sketch graphs where the x axis is a line of symmetry but these are not proper functions as they are not well defined.
eg. x = y^2
B
I know how to do this using algebra. Can you show me using a graph?
It's a cube root not a square root.
To turn the denominator into a rational number you need to multiply by the cube root squared.
Bob
Wow! How on earth did I miss that? Thanks.
Rationalize Denominator
Let cb = cube root
5/[cr{2}]
I multiplied the top and bottom by [cr{2}].
After doing that I ended up with (5•cr{2})/2 as my final answer.
The book's answer is (5•cr{4})/2.
Where did 4 come from?
When the graph of an equation is given, how do I find the intercepts using the graph?
How do I indicate if a given graph is symmetric with respect to the x-axis, y-axis or origin?
NOTE
I HAVE NOT UNDERSTOOD HOW TO UPLOAD PHOTOS TO THIS SITE?
Plot the given points. Then plot the points that are symmetric to the original points with respect to the x-axis, y-axis or the origin.
Given points: (5, 3), (4, -2)
Let me see.
I will not show the plotting of the points because this requires a graph, which means a picture. I have not been able to upload photos to the site.
Symmetric with respect to the x-axis:
(5, 3) ==> (5, -3)
(4, -2) ==> (4, 2)
Symmetric with respect to the y-axis:
(5, 3) ==> (-5, 3)
(4, -2) ==> (-4, -2)
Symmetric with respect to the origin:
(5, 3) ==> (-5, -3)
(4, -2) ==> (-4, 2)
You say?
First part is easy either by calculating the lengths of sides or by using sines/cosines to get the angles.
Second part: Well you could get the coordinates of the midpoints and calculate the lengths of the smaller triangle but that's a lot of work. There's an easier way.
Let's say that an equilateral triangle has vertices A, B and C and that the midpoints are D, E and F with D opposite A, E opposite B and F opposite C
Show that triangles ABC and AEF are similar and hence EF is 0.5 BC.
Similarly show ED and DF are half the side opposite each. Hence EF = DE = DF so DEF is also equilateral.
As this is true for all equilaterals it is true for the one in the above question.
Bob
Ok. I will work this out before Saturday. I have several personal errands to run today. By the way, Suzzane Somers from Three's Company and Step By Step died on Monday October 15 this year one day before her 77th birthday. A beautiful, smart, talented, business lady, actress she was for sure. Love Somers as Chrissy Snow on Three's Company with the late John Ritter and the beautiful Joyce DeWitt. You see, death is no respector of social status.
All agreed.
B
I must say that this problem is not too bad.
Looks ok to me.
B
Very good. By the way, questions posted in this forum are
A. Even number problems which most textbooks don't list the answers to in the back section
B. Problems that I have tried to solve several times but to no avail
C. Problems that I simply don't know how to begin to solve
Thank you. You know, I have a hard time with applications requesting a formula to solve the problem. When a formula or equation is provided no trouble at all.
Determine which of the given points are on the graph of the equation.
Equation: y = x^3 - 2•sqrt{x}
Points: (0, 0); (1, 1); (1, -1)
I think I need to plug and chug here. If I get the same answer on both sides of the given equation, I can then say that the point is on the graph of the given equation.
Let me see.
Point (0, 0).
y = x^3 - 2•sqrt{x}
0 =(0)^3 - 2•sqrt{0}
0 = 0 - 0
0 = 0
Point (0, 0) is the graph of the equation.
How about Point (1, 1)?
y = x^3 - 2•sqrt{x}
1 = (1)^3 - 2•sqrt{1}
1 = 1 - 2(1)
1 DOES NOT = -1.
I say this point is not on the graph of the equation.
How about the final point (1, -1)?
y = x^3 - 2•sqrt{x}
-1 = (1)^3 - 2•sqrt{1}
- 1 = 1 - 2(1)
- 1 = - 1
Yes, this point is on the graph of the equation.
You say?
Verify that the points (0, 0), (a, 0), and (a/2, sqrt{3}a/2) are the vertices of an equilateral triangle. Then show that the midpoint of the three sides are the vertices of a second equilateral triangle.
Any hints?
A hot-air balloon headed due east at a speed of 15 miles per hour and at a constant altitude of 100 feet passes directly over an intersection. Find an expression for the balloon's distance d (measured in feet) from the intersection t minutes later.
Let me see.
60 minutes = 1 hour
Let mph = miles per hour
mph = 15/60 = 1/4.
Distance = (rate)(time) or D = rt.
D = (1/4)t
Let d(t) = Distance apart d (in miles) at the end of t hours.
One mile = 5,280 feet.
So, (1/4)(5,280) = 1320 feet.
Using a right triangle, I came up with the following expression:
d(t) = sqrt{(100)^2 + (1320)^2}.
You say,
Overlay a rectangular coordinate system on a Little League baseball diamond so that the origin is at home plate, the positive x-axis lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base.
A. What are the coordinates of first base, second base, and third base?
B.if the right fielder is located at (180, 20), how far is it from the right fielder to second base?
C. If the center fielder is located at (220, 220), how far is it from the center fielder to third base?
NOTE: THE BASES ARE SEPARATED BY 90 FEET AROUND THE BASEBALL DIAMOND.
Let me see.
For A:
First base coordinates (90, 0).
Second base coordinates (90, 90).
Third base coordinates (0, 90).
You say?
For B:
Let R = distance from right field to second base.
R = sqrt{(180 - 90)^2 + (20 - 90)^2}
Is this correct?
For C:
Let C = distance from center field to third base.
C = sqrt{(220 - 0)^2 + (220 - 90)^2}
You say?
If COVID-19 was not created in a China laboratory, then where did the killer virus come from?
Suzanne Somers of Three's Company and Step By Step dead at 76. She was beautiful, sexy, smart and funny. Thank you, Suzanne Somers for the laughs.
Yes that will work.
B
Ok. Sounds good. I will also work this out on paper. Wish I could upload pictures but without a computer or laptop it is nearly impossible.
Yes that's correct.
B
This is a bit tedious. So, I will work it out on paper. Uploading pictures here is not easy.
That's it!
Bob
It certainly feels good to get the right answer. It creates a sense of personal accomplishment.
You can 'read' this as if maths is a language.
Curly brackets means the set
with the points of the form (x,3)
: such that
x is a member of the real numbers.
Put the underlined bits together and you get
the set with the points of the form (x,3) such that x is a member of the real numbers.
Bob
This was helpful. I thank you for the translation.
This interactive tool of MIF's may help visualise what's going on: Interactive Cartesian Coordinates
It has several options you can choose from (eg, regular or irregular shapes, choice of the number of sides, entry of custom coordinates).
However, it will only display one shape at a time, and, unlike Geogebra, it lacks mathematical precision...but it's accurate enough to give a pretty good idea.
Thank you for the link. I truly appreciate your reply.
If we take (0,4) and (0,0) as the two points defining the base then the midpoint is (0,2) and the horizontal line y=2 is the line of symmetry that goes through the vertex. So it must have its y coordinate 2.
If you use Pythag. on the half triangle made by splitting the equilateral triangle in two along that line (y=2) you can determine the required x coordinate(s). There will be two answers as the triangle may 'sit' left of the y axis or right of it.
Bob
Thank you. I will work this out on paper.
An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral triangle are (0, 4) and (0, 0), find the third vertex. How many of these triangles are possible?
Seeking the needed steps here.
The medians of a triangle are the line segments from each vertex to the midpoint of the opposite side. Find the lengths of the medians of triangle ABC with vertices at A = (0, 0), B = (6, 0), and C = (4, 4).
Let me see.
The author gave no sample question for this problem.
I think this is what I must do:
1. Find the midpoint of line segments AB, BC, and AC.
2. I then must use the distance formula for points to find the following:
A. Distance form point A to the midpoint of BC.
B. Distance from point B to the midpoint AC.
C. Distance from point C to the midpoint AB.
You say?
Plot the points A = (-1, 8) and M = (2, 3) on the xy-plane. If M is the midpoint of the line segment AB, find the coordinates of B.
Let me see.
Let A = (-1, 8) = (x_1, y_1)
We are given the midpoint coordinates to be (2, 3).
Let B = (x_2, y_2)
Here is my set up to find x_2:
2 = [(-1 + x_2)/2]
Here is my set up to find y_2:
3 = [(8 + y_2)/2]
Is this correct this far?