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The current time is 12 noon CST. What time (CST) will it be 12,997 hours from now?
The question looks easy enough but it's tricky to me.
I know there are 24 hours in a day. So, I decided to divide 12,997 by 24 hours, which yield the decimal number 541.54166666667.
Of course, this decimal number tells me nothing.
A. How do I find the correct answer?
B. If the quotient found leads to the right answer, can you show me how?
A rational number is defined as the quotient of two integers. When written as a decimal, the decimal will either repeat or terminate. By looking at the denominator of the rational number, there is a way to tell in advance whether it's decimal representation will repeat or terminate. What about the denominator of a rational number indicates that its decimal representation will repeat or terminate?
Help me! I newbie!
Help you with what? We need a specific question with clear instructions. So, post your question(s) and wait for someone to reply.
Copy and paste thus url to see question.
https://imgur.com/a/ZSarUyQ
Copy and paste the link provided to see image.
https://imgur.com/a/qtBeMMK
amnkb wrote:"odd" means it matches on both sides of diagonal y = x
**
That's not the definition I have always used.
https://www.mathsisfun.com/algebra/func … -even.html
The url in post 3 worked when I made my post but not now. Maybe the poster removed it from imgur ???
There were two graphs. The first was y = x^2 or something similar. The second was, I think, y = e^x or some thing similar. There were no scales so the exact function could not be determined but enough to show the symmetry or not.
Bob
ps ** If a graph of a function is reflected in the line y=x this is equivalent to interchanging y and x and the resulting graph shows the inverse function.
eg. Compare y = 2x + 3 and y = (x-3)/2
Hi Bob.
Yes, I removed the photo from imgur. I will post several questions later showing my work or at least effort. I need all the practice I can get with imgur.
I am not savvy in terms of computers, laptops, tablets, and cell phones. I just know internet basics. Again, I deleted all imgur uploads yesterday. I will keep my uploaded stuff there moving forward.
harpazo1965 wrote:Ok. Thanks. Now, I am having a hard uploading images to the site.
Here is a question about Even, Odd ,Neither.
The image is found here:
https://imgur.com/a/hQDSFH4
putting in img tags doesnt work
a. i think the 1st fcn is y = 1/x (imnshp, when they don't give you fcn its kind of unfair to tell you to find stuff out about it)
"even" means it matches on both sides of the y axis
"odd" means it matches on both sides of diagonal y = x
this fcn is odd because you can fold graph on diagonal and teh 2 halfs will match
domain is x not equal to 0 (im guessing from the graph; it looks like there is a vertical asymptiote at x=0)
range is y not equal to 0 (im guessing from the graphl it looks like there is a horizontal asymptote at y=0)
b. it looks like maybe an exponential, like y=e^x ?
if you fold the graph on the y axis the halfs wont match so its not even
if you fold on the diagonal the halfs wont match so its not odd
domain looks like all x
i dont know how your supposed to know the range from the picture; the graph could cross the axis off to the left
but if you guess that the graph close to the center tells you everything you need to know then it looks like graph is always above x axis
Bob saw my photo. This is what matters.
A function is said to be even if it is symmetrical in the y axis. Here are some examples:
y = x^2
y = x^4
y = x^6 + x^4 - x^2
From these examples you can probably see why these are called even functions.
To test algebraically replace every x in the function with -x. If you get the same function back, then it's even.
example:
y = x^2 Replace x with -x .... ynew = (-x)^2 = x^2 = y. So y is even.
In a similar way odd functions have rotational symmetry around the origin order 2. ie. If you rotate the graph 180 around (0,0) and the graph looks the same then it is a odd function.
example:
y = x^3
But most likely, a function is neither The second example looks like y = e^x and this is neither odd nor even.
If you track left on the graph, as x tends to minus infinity y tends to zero. There are no negative y values. As you track right, as x tends to + infinity y tends to + infinity. So the range is (0, + infinity). The round brackets indicate that the endpoints are never achieved.
Bob
Did you see the photo? How did you see the photo? I cannot figure out how to upload photos. As long as you can see my URL photo, we are in business. I like your explanation about even, odd and neither functions. Thank you.
Ok. Thanks. Now, I am having a hard uploading images to the site.
Here is a question about Even, Odd ,Neither.
The image is found here:
https://imgur.com/a/hQDSFH4
Factor completely.
5 + 16x - 16x^2
Let me see.
5(-16) = -80
What two numbers when multiplied yield -80 but when added produce 16?
How about 20 and -4?
20(-4) = -80
20 + (-4) = 16
Yes, 20 and -4 will do.
Put it all together to force factor by grouping.
5 + 20x - 4x - 16x^2
Factor by grouping.
5 + 20x = 5(1 + 4x)...Group A
-4x - 16x^2 = -4x(1 + 4x)...Group B
Put both groups together.
5(1 + 4x) - 4x(1 + 4x)
My answer: (5 - 4x)(1 + 4x)
Book's answer: -(4x - 5)(4x + 1)
What did I do wrong?
harpazo1965 wrote:Bob wrote:The two expressions have p^2 in it and 2ap if you put a = 7
So add a^2 = 49 and you have a perfect square.
Let me see.
(14/2)^2 = (7)^2 = 49
The missing number that completes the square is 49.
You say?
yes that is what he said
Ok. Cool.
Thank you.
We're in luck here because (x^3)^2 = x^6
So substitute Y = x^3 and you'll get a quadratic in Y.
Bob
Let me see.
Rewrite x^6 as (x^3)^2.
Let u = x^3
u^2 + 2u + 1
Factor.
(u + 1)(u + 1)
Back-substitute for u.
(x^3 + 1)(x^3 +1)
The sum of cubes tells me that (x^3 + 1) is the same thing as
(x + 1)(x^2 - x + 1).
My answer is (x + 1)(x^2 - x + 1)^2.
Textbook answer is (x + 1)^2(x^2 - x + 1)^2.
Who is right and why?
Note that
The two expressions have p^2 in it and 2ap if you put a = 7
So add a^2 = 49 and you have a perfect square.
As a general rule find half the middle coefficient (14/2 = 7) and square it (49).
Bob
Let me see.
(14/2)^2 = (7)^2 = 49
The missing number that completes the square is 49.
You say?
Bob,
Thanks again. I will try a few on my own from the textbook.
I really don't understand how to upload pictures to this site.
Can you share the steps with me? Eventually, I will need to upload geometric and trigonometric pictures. Keep on mind that I don't have a computer or laptop. Can I upload pictures using my android cell phone?
Also, I will not reveal my name in an open forum. I will use my initials instead.
GF
Bob,
Thank you. You said we're in luck in this case because (x^3)^2 = x^6.
What if the question is not so obvious? What then?
Factor completely. If the polynomial cannot be factored, say it is prime.
x^6 + 2x^3 + 1
Determine the number that should be added to complete the square for the given expression.
p^2 + 14p
Thank you.
harpazo1965 wrote:I am thinking of a number. It lies between 1 and 10; it's square is rational and also lies between 1 and 10. The number is larger than pi. Correct to 2 decimal places (that is, truncated to two decimal places) name the number.
ne # between these works
infinitely many others do 2
Thank you. Interesting question.
harpazo1965 wrote:Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.
NOTE: Looking for the set up only.
e_jane_aran wrote:I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.
If you add two of the above expressions, you get the third of the above expressions. Which pair works?
harpazo1965 wrote:If I add a^2 [and] c^2, the middle term cancels out. This does not leave me with much to play with.
If I do that, I am left with m^4 + n^4 twice. In other words, I get
2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.so maybe try a different pair of terms? (there r 2 other pairs; 1 of them works)
If I add a^2 + b^2 I get c^2. So, the addition of the two legs of this triangle
yields the value of the hypotenuse. You say?
harpazo1965 wrote:Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.
NOTE: Looking for the set up only.
I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.
If you add two of the above expressions, you get the third of the above expressions. Which pair works?
If I add a^2 + c^2, the middle term cancels out. This does not leave me with much to play with.
If I do that, I am left with m^4 + n^4 twice. In other words, I get
2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.
Ok. No problem. Let's get back to math.
You need to show that two of these when squared and added make the same result as the third when squared.
But which to choose? I can see that c is bigger than a. So I suggest working out c^2 and write that down. Then get an expression for a^2 + b^2. If this comes to the same as c^2 then you are done.
Bob
Thank you, Bob.
College Algebra
Section R.3
Suppose that m and n are positive integers with m > n.
If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.
Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.
NOTE: Looking for the set up only.