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Firstly, multiply every term by x.(x+10)
Then, cancel the x in the first term and the (x+10) in the second to leave
Deon588,
but don't use decimal points as above to multiply. Even if they *were* to
mean multiplication, then they would be redundant. Use grouping
symbols instead.
Just as 3.4 equals 3 and four-tenths, it does not equal
And you would not type 200.x to be the equivalent of 200 multiplied by x,
because 200x means 200 multiplied by x.
Then, cancel the x in the first term and the (x+10) in the second to leave
Sorry for not being totally clear annonimnystify.
The question is write the domain of f(x) and solve the equation.
Zetafunc do you mean break it up like this log_2(x-4)=log_4((x-8)
sorry for writing it like this, I don't know much about LaTeX yet...
If this is the way to go how would the change of base step look?
should it be log_2/log_2(x-8)? Thanks a lot
Deon588,
please make some breaks in your lines such as in this amended
quote box above.
Can someone point me in the right direction to simplify this equation.
1/x+1 - 1/2x+1 = 1/6
thanks
planetdom.
*again*, place grouping symbols around the
appropriate denominator forms. (The last
denominator would not have them as they
are unnecessary.)
Practice typing what I have been telling you.
1/(x + 1) - 1/(2x + 1) = 1/6
or
------------------------------------------------------------
Otherwise, your original equation is equivalent to:
Hi,
I was hoping someone could explain why
3/√x + √x = 3/√x + √x.√x/√X
when then simplifies further to
3+x/√x
many thanks
planetdomi,
you need grouping symbols. For example,
you must enclose the numerator for your final expression.
The product of that fraction in your first line worked out
because of a fluke. Also, use the same lower case letter
throughout the exercise for the same variable.
Look at this:
3/√x + √x =
3√x + (√x/1)(√x/√x) =
3/√x + x/√x =
(3 + x)/√x
hi i did number 10
simplify :
i got
* * * Note: Spread out each new expression on one line with one equals sign per line.
-----------------------------------------------------------------------------------------------------------
oooh so u substituted wow that was easy thanx
i did 11 am working on 12
#11-simplify
* * * Follow this type of amendment:
Bob
Strong advice for everyone:
Do not mix the "times sign" with the algebraic operations/expressions.
Keep the "times sign" within arithmetic.
Keep the "times sign" out of algebra.
Instead, use a multiplication dot or grouping symbols for the multiplication.
------------------------------------------------------------------------------------------
zee-f,
when solving any equation where the variable is in a denominator,
make sure you omit any (candidate) solutions that would cause
one (or more) denominators to equal zero, because those would
make the value of those algebraic fractions undefined.
Is this convergent or divergent?
Please look at this amended problem. (All variables are the same now.)
-----------------------------------------------------------------------
-----------------------------------------------------------------------
"Since;
Has no solutions, ..."
There was a misunderstanding here by me,
because you didn't put this all on one line**,
and you had "Has" in the middle of a
sentence instead of "has." It appears to
be two sentences, despite "Since;[,]"
being at the beginning of the sentence.
** It is just about the visual aspect of it not being on
one line, not the correctness of it.
*** Edit:
My amended question is still open.
Hi;
Since;
1)
2) Has no solutions,
the summand is a constant
and an infinite sum of constants does not converge.
1)
You meant the *limit of that expression as x --> oo*, correct?
Can you justify this?
2)
So, I don't see your first part of your message as being sufficient
to justify that the sum is divergent (not convergent).
Is this convergent or divergent?
Please look at this amended pronblem. (All variables are the same now.)
hey i was wondering if i did these right
For problems 1-3 (?), solve the given rational equation.
#1 I choose D --> x = 8/9
#1-
A) x = 9/8
B) x = 1
C) x = 2
D) x = 8/9
E) x = 2x
F) x = 4-------------------------------------------------------------
#2 I choose B --> k = 20/3
#2-
A) k = 2
B) k = 20/3
C) k = 3
D) k = 3/20
E) k = 7
F) k = -1-----------------------------------------------------------
#3 I choose F --> n = 11/4
#3
A) n = 1/2
B) n = 11/2
C) n = 1/5
D) n = 1/4
E) n = 4/11
F) n = 11/4
------------------------------------------------------------
for one thing, put in horizontal spaces and vertical
spaces for readability. And, taken literally, "Ax,...,
Ak,..., and An,... are meaningless here.
For another thing, I have *already* told you that you
don't type "[/math]" and "[math]" in the middle of
your equations, because the equals signs do not
expressed (in Latex.)
All of your *three* answers are correct.
***Edit:
In my opinion, your edit has helped very much.
Regarding the color used in post #2:
Would you be able to use a color other than the one you used
to reply to Macy, because the contrast is relatively harsh against the
normal color background?
If not, thank you anyway.
#6-
#7-
#8-
#9-
#10-
zee-f,
don't have [/math] \div [math] in the middle of your lines.
Type [1st expression] \div [2nd expression]
Click onto my quote button on the bottom right of my post
to see how I amended yours.
Hi Maiya;
The above should be more complete.
1) Prove: if n<0 and |m|= m-n/m+n , -1<m<0 or m>1
2) Given m = 4-x/3, n = x+3/4, p = 2-3x/5 , and m>n>p
Prove that the possible value of x is {-7/17<x<1}
fromwodehouse,
because of the Order of Operations,
you must use grouping symbols for these:
** including square roots of certain integers
New Problem:
You are given an integer t ( 0 to 9 ). You can use decimal points, multiplication, addition, division, subtraction, parentheses and exponentiation only. How close can you get to Euler's number using t only? You can use t as many times as you want.
A says) I got it using 3 t"s.
B says) Nonsense! Not with 3 you didn't.
C says) 4 t's is about right.
D says) Yep!
E says) 3 is definitely not the best.What is t and what is your expression?
hi
i think rya's soluton is better because it doesn't use digits other than 3 3s
anonimnystefy,
his solution does only use three 3s and no other digits.
That "100" is not part of it as a base; it is following the equals sign as
what his expression is equal to.
Let me know if that is what you were referring to, please.
Form expressions (in three problems) that are equivalent to 100.
For each of these three problems, you must use all of the three
digits and no other digits.
Also, you are allowed only these:
concatenation of digits
decimal point
addition, subtraction, division**, multiplication signs
(and/or use grouping symbols for products)
grouping symbols such as parentheses, brackets, and braces
factorial symbol, and possible nesting of these is permitted
square root symbol, and nesting of these is also permitted
exponentiation
floor and ceiling functions
--------------------------------------------------------------------------
--------------------------------------------------------------------------
A) three 1s
------------------------
------------------------
B) three 2s
------------------------
------------------------
C) three 3s
------------------------
------------------------
.
The new shape is an exact copy but twice as big.
Bob
The following is a case where the new shape is twice as big as the original shape:
Suppose you have a rectangle with dimensions of
If each dimension is multiplied by
then the new area is doubled. That is, you have a shapeHowever, (again i'm repeating myself from my other post), if the width *and* the length are doubled, then the shape
will be four times as big. This means the area has been quadrupled.
My Mom and I have talked about this doubling or halving of the size
of a photograph recently. It is interesting because there is no
consensus for regular non-math people on what half or double the
size of something really even means.
Take it in context.
If you double the size of a line segment, its length is doubled.
A circle, rectangle, or a photograph's size is its area. For
a rectangle, only one dimension (the width or the length)
need be doubled to double the size of the rectangle.
However, if each of the two dimensions of the rectangle is
doubled, its size is quadrupled.
A sphere and a rectangular box's size, for instance,
are their volumes. If exactly one of the dimensions
of the box (width, length, or height) is doubled, then
the size of that box (its volume) is doubled.
However, if each of the three dimensions (of the box)
is doubled, then the volume of the new box is eight
times the volume of the original box's volume.
Regarding the Moon, it is approximately spherical.
Its size is its volume.
However, when someone speaks of the apparent
size of a full Moon looking larger than at other times,
they are looking at a shape which is circular (an area).
Then, they should not be meaning the actual Moon's
radius or circumference (linear dimensions), and they
should not be meaning the actual Moon's volume,
which is cubic.
When the full Moon's size is said to appear larger,
then the area of the circular side showing should
be what is meant.
** The Moon may appear larger to some people when
viewed from the horizon versus elsewhere in the sky
from visual cues near its image on the horizon,
versus being well above the horizon.
"...over and under the point 0 of the
X acces.
If you go towards the negative or positive part of the X access..."
1 + 11 + 11 ≠1111
That's an inequation. That's not an equation.
mathsyperson,
yours is an inequality, so it's not allowed either.
If we move one of the sticks from the group of 3(III) and place it with the single stick (I):
II + II + II = IIII
we have 6 sticks on the left side , while we have 4 sticks on the other
So ..
polylog solution is wrongLet me suggest this idea:
I + (II) + III = IIII
I + (+) + III = IIIIConverting the (II) into (+) , will give 1 + 3 = 4
My calculator gives an error for
1 + + + 3, so I am voting against that.
bobbym,
that works in the Roman numeral system.
IIII also equals IV.
You have a list of prime numbers in the far left column,
where they are being used for analysis/calculations.
It looks as though you have to have those prime numbers
in advance to get a longer list of greater prime numbers.