Accounting does not require advanced algebra or trigonometry, you may do fine with basic knowledge in Maths. What is hard about accounting is that it involves analyzing, reporting which requires constant attention. Instead of maths, maybe you should think if you are into economics and finance...
Is accounting hard? I am not too good in math.I had B in algebra and C in trig%26amp; trig?
i did business and finance for three years algebra is only used in business statistics if this is what you are intersted in. it involves plotting a busineess' future as in future earnings whether they would make a profit or a loss.
the finance side would be accounting and it does get complicated doing petty cash is simple but it gets harder when doing profit and loss accounts on triple ledger. but i suppose it would be easier nowadays with all the finance programs you can get such as sage. you also need to know how to use spreadsheets as well
Reply:I don't think it is hard. I took a Business course and to my surprise "basic accounting principles" were one of the classes. I was kind of nervous about it. To make a long story short it was my favorite class. It was fun to learn, got a 4.0. Im sure you would be fine.
sweet pea
Tuesday, July 14, 2009
What value of c would give a system with no solution? Math help!!?
x + 2y = 6
cx - 4y =8
Please show all the stepas, to this equation. thanks
What value of c would give a system with no solution? Math help!!?
multiply 1st equation by -2 to match up y coefficients:
-2x - 4y = -12
cx - 4y = 8
if c = -2 there will be no solution, since
-2x - 4y cannot equal both -12 and 8
cx - 4y =8
Please show all the stepas, to this equation. thanks
What value of c would give a system with no solution? Math help!!?
multiply 1st equation by -2 to match up y coefficients:
-2x - 4y = -12
cx - 4y = 8
if c = -2 there will be no solution, since
-2x - 4y cannot equal both -12 and 8
How come some students in my class are in algebra 2 and goes to the high school for math? is there any way i c
please help me
How come some students in my class are in algebra 2 and goes to the high school for math? is there any way i c
They probably tested for higher placement in math. Ask your teacher this question, and express your interest in doing the same thing.
Reply:I almost did that but then I didn't cuz I enrolled into the Challenge program,
So here's what you do: Go to your school counsler and take The Test to see if you qualify or not
How come some students in my class are in algebra 2 and goes to the high school for math? is there any way i c
They probably tested for higher placement in math. Ask your teacher this question, and express your interest in doing the same thing.
Reply:I almost did that but then I didn't cuz I enrolled into the Challenge program,
So here's what you do: Go to your school counsler and take The Test to see if you qualify or not
Can you help me out with this math problem? Show that [x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]=1?
please do not solve this using logarithm method; my teacher told me specifically only to solve this arithmetically
Can you help me out with this math problem? Show that [x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]=1?
x^a(b-c) = x^(ab - ac) = x^(ab)/x^(ac)
x^b(a-c) = x^(ab - bc) = x^(ab)/x^(bc)
Now, [x^a(b-c)]/[x^b(a-c)] =
= [x^(ab)/x^(ac)] / [x^(ab)/x^(bc)] =
= [1/x^(ac)]/[1/x^(bc)] =
= 1/x^(ac) * x^(bc) =
= x^(bc)/x^(ac) =
= [(x^b)^c]/[(x^a)^c] =
= [(x^b/x^a)^c]
QED
Reply:x^[a(b-c)]/[x^b(a-c)]=x^[ab-ac-ab+bc]=x^...
=[x^b/x^a)^c].
this easily leads to the final result desired.
Reply:[x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]
[x^ab-ac /x^ab-bc ] divided by [(x^b-a)^c]
x^ab-ac -ab +bc divided by x^c(b-a)
x^c(b-a) divided by x^c(b-a)
=1
Reply:Let's start with the first half of the problem:
[x^a(b-c)]/[x^b(a-c)]
You can bring the denominator to the top by making the exponent negative:
[x^a(b-c)] * {x^-[b(a-c)]}
When you do all the distributions, you get:
(x^ab-ac) * (x^-ba+bc) or
(x^ab-ac) * (x^-ab+bc)
When you multiply two (same) bases that have exponents, you simply add the exponents, keeping the base:
x^bc-ac
(The ab terms got canceled)
Now we can apply the same rules to the second part of the problem:
[(x^b/x^a)^c]
Remember that this whole term was in the denominator, so we can bring c to the top by reversing its sign:
[(x^b/x^a)^-c]
And we can do the same for a:
[(x^b * x^-a)^-c]=
Now when you have a power raised to another power, you simply multiply the exponents:
(x^-bc * x^ac)
Adding these exponents, we get:
(x^-bc+ac)
Now our full problem looks like:
x^(bc-ac) * x^(-bc+ac) = 1
Now when you add those exponents, you get zero:
x^0 = 1
And anything raised to the zero power is 1:
1 = 1
Can you help me out with this math problem? Show that [x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]=1?
x^a(b-c) = x^(ab - ac) = x^(ab)/x^(ac)
x^b(a-c) = x^(ab - bc) = x^(ab)/x^(bc)
Now, [x^a(b-c)]/[x^b(a-c)] =
= [x^(ab)/x^(ac)] / [x^(ab)/x^(bc)] =
= [1/x^(ac)]/[1/x^(bc)] =
= 1/x^(ac) * x^(bc) =
= x^(bc)/x^(ac) =
= [(x^b)^c]/[(x^a)^c] =
= [(x^b/x^a)^c]
QED
Reply:x^[a(b-c)]/[x^b(a-c)]=x^[ab-ac-ab+bc]=x^...
=[x^b/x^a)^c].
this easily leads to the final result desired.
Reply:[x^a(b-c)]/[x^b(a-c)] divided by [(x^b/x^a)^c]
[x^ab-ac /x^ab-bc ] divided by [(x^b-a)^c]
x^ab-ac -ab +bc divided by x^c(b-a)
x^c(b-a) divided by x^c(b-a)
=1
Reply:Let's start with the first half of the problem:
[x^a(b-c)]/[x^b(a-c)]
You can bring the denominator to the top by making the exponent negative:
[x^a(b-c)] * {x^-[b(a-c)]}
When you do all the distributions, you get:
(x^ab-ac) * (x^-ba+bc) or
(x^ab-ac) * (x^-ab+bc)
When you multiply two (same) bases that have exponents, you simply add the exponents, keeping the base:
x^bc-ac
(The ab terms got canceled)
Now we can apply the same rules to the second part of the problem:
[(x^b/x^a)^c]
Remember that this whole term was in the denominator, so we can bring c to the top by reversing its sign:
[(x^b/x^a)^-c]
And we can do the same for a:
[(x^b * x^-a)^-c]=
Now when you have a power raised to another power, you simply multiply the exponents:
(x^-bc * x^ac)
Adding these exponents, we get:
(x^-bc+ac)
Now our full problem looks like:
x^(bc-ac) * x^(-bc+ac) = 1
Now when you add those exponents, you get zero:
x^0 = 1
And anything raised to the zero power is 1:
1 = 1
How do you call those math/logic problems where A =B, B = C and C = A?
Do you have any examples?
How do you call those math/logic problems where A =B, B = C and C = A?
It's called transitivity
example:
{x =y and y =z } ----%26gt; x =z
Reply:this is call ed the transitive law
Reply:Just like substitution if y= 3x+5 and y=4x+6
since we know y = 3x+5 we can substitute that into the other equation for y:
3x+5 = 4x+6
Now apply this same logic to your example
but instead:
y will be B
3x+5 will be A
4x+6 will be C
so instead of:
y= 3x+5 we have B=A
instead of:
y=4x+6 we have B=C
since B=A and B also =C
we can say that A and C (A=C) must be the same as well.
Hope that was helpful! :)
rose
How do you call those math/logic problems where A =B, B = C and C = A?
It's called transitivity
example:
{x =y and y =z } ----%26gt; x =z
Reply:this is call ed the transitive law
Reply:Just like substitution if y= 3x+5 and y=4x+6
since we know y = 3x+5 we can substitute that into the other equation for y:
3x+5 = 4x+6
Now apply this same logic to your example
but instead:
y will be B
3x+5 will be A
4x+6 will be C
so instead of:
y= 3x+5 we have B=A
instead of:
y=4x+6 we have B=C
since B=A and B also =C
we can say that A and C (A=C) must be the same as well.
Hope that was helpful! :)
rose
What kind of math is this A SQUARE + B SQUARE = C SQUARE?
It's loud math.
But A^2 + B^2 = C^2 is the pythagorean theorem.
What kind of math is this A SQUARE + B SQUARE = C SQUARE?
It's the Pythagorean theorem, used to determine the length of a hypotenuse of a right triangle. It's part of Euclidean geometry.
Reply:It;'s algebra, but there is no way to solve it without either values for two of the variables.
Reply:that is the pythagorean theorem.
it is used in finding the measure of the sides of a right triangle.
for example:
3, 4 and 5
3^2 + 4^2 = 5^2
9 + 16 = 25
25 = 25
Reply:That could be geometry or trigonometry. Or even part of a calculus problem.
Reply:Well, it's the Pythagorean theorem, a fact about geometry.
In the days of Pythagoras, geometry was used to prove results in all kinds of mathematics, or at least as much mathematics was then known; "geometry" MEANT "mathematics". But now that's not the case.
On the other hand, that formula is central to the mathematical definition of "distance", and hence lies at the heart of various areas of mathematics, from statistics to "linear algebra" to "differential geometry". It's also pretty central to trigonometry.
Reply:That is the pythagoras' theorem. We use it to find the length of a side of a right angled triangle.(i.e 1 angle is 90 degree)
The longest side is called the hypotenuse. In this case it is C.
We can use this to find a side of two sides of a right angled triangle is given.
But A^2 + B^2 = C^2 is the pythagorean theorem.
What kind of math is this A SQUARE + B SQUARE = C SQUARE?
It's the Pythagorean theorem, used to determine the length of a hypotenuse of a right triangle. It's part of Euclidean geometry.
Reply:It;'s algebra, but there is no way to solve it without either values for two of the variables.
Reply:that is the pythagorean theorem.
it is used in finding the measure of the sides of a right triangle.
for example:
3, 4 and 5
3^2 + 4^2 = 5^2
9 + 16 = 25
25 = 25
Reply:That could be geometry or trigonometry. Or even part of a calculus problem.
Reply:Well, it's the Pythagorean theorem, a fact about geometry.
In the days of Pythagoras, geometry was used to prove results in all kinds of mathematics, or at least as much mathematics was then known; "geometry" MEANT "mathematics". But now that's not the case.
On the other hand, that formula is central to the mathematical definition of "distance", and hence lies at the heart of various areas of mathematics, from statistics to "linear algebra" to "differential geometry". It's also pretty central to trigonometry.
Reply:That is the pythagoras' theorem. We use it to find the length of a side of a right angled triangle.(i.e 1 angle is 90 degree)
The longest side is called the hypotenuse. In this case it is C.
We can use this to find a side of two sides of a right angled triangle is given.
What is this math equation called: a(squared)+b(squared)=c(square...
it's the Pythagorean Theorem
In any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.
What is this math equation called: a(squared)+b(squared)=c(square...
The Pythagorean Theorem for right angle triangles.
Reply:its the pythagorean theorem
Reply:The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.
Pythagorean theorem, after Pythagoras, an ancient Greek geometer.
If one leg making the right angle is 5, and the other leg making the right angle is 4, then the remaining side of the triangle, the hypotenuse, is [5x5]+[4x4]=41.
Reply:Quadratic...? I think...
Reply:pythagorean theorem
Reply:pythagorean theorem
Reply:It is the Pythagorean Theorem.
A(square)+B(square)=C(square)
Remember, it is for right triangles.
Reply:Pythagorean Theorem or if you like to get technical the right triangle theorem.
Reply:i think you're talking about the pythagorean theorem
Reply:pythagoreen theorom
Reply:From Wikipedia, the free encyclopedia
The Pythagorean theorem: The sum of the areas of the two squares on the legs (blue and red) equals the area of the square on the hypotenuse (purple).In mathematics, the Pythagorean theorem or Pythagoras's theorem is a relation in Euclidean geometry between the three sides of a right triangle. In the West, the theorem is named after the Greek mathematician Pythagoras, who is credited with the first abstract proof.
The theorem is as follows:
In any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.
If we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the following equation:
This equation provides a simple relation between the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. A generalization of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them.
This theorem may have a greater variety of known proofs than any other. The Pythagorean Proposition, a book published in 1940, contains 370 different proofs of Pythagoras's theorem, including one by an American President James Garfield.
The converse of the theorem is also true:
For any three positive numbers a, b, and c such that a2 + b2 = c2, there exists a triangle with sides a, b and c, and every such triangle has a right angle between the sides of lengths a and b.
Reply:pythagorean theorem
Reply:Pythagorean Theorem
Reply:Pythagorean theorem
In any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.
What is this math equation called: a(squared)+b(squared)=c(square...
The Pythagorean Theorem for right angle triangles.
Reply:its the pythagorean theorem
Reply:The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.
Pythagorean theorem, after Pythagoras, an ancient Greek geometer.
If one leg making the right angle is 5, and the other leg making the right angle is 4, then the remaining side of the triangle, the hypotenuse, is [5x5]+[4x4]=41.
Reply:Quadratic...? I think...
Reply:pythagorean theorem
Reply:pythagorean theorem
Reply:It is the Pythagorean Theorem.
A(square)+B(square)=C(square)
Remember, it is for right triangles.
Reply:Pythagorean Theorem or if you like to get technical the right triangle theorem.
Reply:i think you're talking about the pythagorean theorem
Reply:pythagoreen theorom
Reply:From Wikipedia, the free encyclopedia
The Pythagorean theorem: The sum of the areas of the two squares on the legs (blue and red) equals the area of the square on the hypotenuse (purple).In mathematics, the Pythagorean theorem or Pythagoras's theorem is a relation in Euclidean geometry between the three sides of a right triangle. In the West, the theorem is named after the Greek mathematician Pythagoras, who is credited with the first abstract proof.
The theorem is as follows:
In any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.
If we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the following equation:
This equation provides a simple relation between the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. A generalization of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them.
This theorem may have a greater variety of known proofs than any other. The Pythagorean Proposition, a book published in 1940, contains 370 different proofs of Pythagoras's theorem, including one by an American President James Garfield.
The converse of the theorem is also true:
For any three positive numbers a, b, and c such that a2 + b2 = c2, there exists a triangle with sides a, b and c, and every such triangle has a right angle between the sides of lengths a and b.
Reply:pythagorean theorem
Reply:Pythagorean Theorem
Reply:Pythagorean theorem
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