Featured 1 month ago

A fraction's structure is such that we have:

You can not

Multiply by 1 and you do not change the value of something. However, 1 comes in many forms so you change the way something looks without changing its value.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now that the bottom numbers or size indicators (denominators) are the same we can just compare the top numbers or counts (numerators).

Mathematically you say: multiply both sides by 15#

So we now have:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If we divide the left hand side by the right hand side we will get 1 if they are the same.

Featured 1 month ago

Given:

#x^3+y^3#

Note that if

#x^3+y^3=(x+y)(x^2-xy+y^2)#

We can calculate the discriminant for the remaining homogeneous quadratic in

#x^2-xy+y^2#

is in standard form:

#ax^2+bxy+cy^2#

with

This has discriminant

#Delta = b^2-4ac = color(blue)(1)^2-4(color(blue)(1))(color(blue)(1)) = -3#

Since

We can factor it with complex coefficients by completing the square and using

#x^2-xy+y^2 = (x-1/2y)^2+3/4y^2#

#color(white)(x^2-xy+y^2) = (x-1/2y)^2+(sqrt(3)/2 y)^2#

#color(white)(x^2-xy+y^2) = (x-1/2y)^2-(sqrt(3)/2 y i)^2#

#color(white)(x^2-xy+y^2) = ((x-1/2y)-sqrt(3)/2i y)((x-1/2y)+sqrt(3)/2i y)#

#color(white)(x^2-xy+y^2) = (x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)#

So:

#x^3+y^3 = (x+y)(x^2-xy+y^2)#

#color(white)(x^3+y^3) = (x+y)(x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)#

Featured 1 month ago

See a solution process below:

First, let's call the number Jaden started with:

**He adds 3 to the number**

**Then doubles the result**

**Finally, he subtracts 7**

** He ended with 9**

We can solve this equation by first adding

Next, divide each side of the equation by

Now, subtract

Jaden started with the number:

**He adds 3 to the number**

**Then doubles the result**

**Finally, he subtracts 7**

Featured 1 month ago

Answer:-#" "color(red)(1/n)# and#color(red)(1/m#

If the roots of an equation

#color(red)(ax^2+bx+c=0# is#color(blue)(alpha,beta# , then we can write as per rule that

#color(red)(alpha+beta)=-b/a# #color(red)(alpha cdot beta)=c/a# As per given condition, we can write that

#color(red)(m+n)=-b/a# #color(red)(m cdot n)=c/a# We will determine some values now for further use.

#color(red)(-b/c)=(-b/a)/(c/a)=(m+n)/(m cdot n)# #color(red)(a/c)=1/(m cdot n# If the roots of the equation

#color(red)(cx^2+bx+a=0# is#color(blue)(alpha,beta# , then,

#color(red)(alpha+beta)=-b/c=(m+n)/(m cdot n)" "...(1)# #color(red)(alpha cdot beta)=a/c=1/(m cdot n#

#color(red)(alpha-beta)#

#=sqrt((alpha+beta)^2-4 cdot alpha cdot beta)#

#=sqrt(((m+n)/(m cdot n))^2-4 /(m cdot n))#

#=(m-n)/(m cdot n)" "...(2)#

- From
#(1)" & " (2)# ,

#color(red)(alpha)=((m+n)/(m cdot n)+(m-n)/(m cdot n))/2=1/n# #color(red)(beta)=((m+n)/(m cdot n)-(m-n)/(m cdot n))/2=1/m# Hope this helps....

Thank you...

Featured 1 month ago

**Questions:**

How do you find the value of the combination

**"seven choose five"**.

Or, it can be read as, **"choose five from seven"**.

The calculation can be used when you have seven objects and want to know how many ways you can make a collection of five objects, without considering in which order they are to be chosen.

For example, you may want to choose a basketball team of 5 people from a group of 7, to start the game and the players are of approximately equal ability.

Featured 1 month ago

2 real solutions

You can use the discriminant to find how many and what kind of solutions this quadratic equation has.

Quadratic equation form:

Discriminant:

Plug 2, 1, and -1 in for a, b, and c (and evaluate):

Negative discriminants indicate that the quadratic function has 2 imaginary (involving

Discriminants of 0 indicate that the quadratic function has 1 real solution. The quadratic function can be factored into the perfect square of something (such as

Featured 3 weeks ago

Letters!

The biggest change in mathematics when entering algebra is the use of ** variables**. Variables are able to represent more general ideas in mathematics, which can allow for problems to be solved more easily.

Let's look at an example of how algebra is useful to us. Consider a scenario where you have apples. You want the following 3 conditions to be true.

[1] You want to be able to give half of your apples to your first sibling.

[2] You also want your second sibling to have

[3] The total number of apples you want to buy at the store is

**How many apples does each person get?**

The "not algebra" approach would be to randomly guess how many apples each person gets and check that all 3 conditions are met. This might work, but for complicated problems it becomes difficult. Algebra helps us out by allowing us to logically reach the solution.

The algebraic approach is the write down a set of equations using variables. Let's do that. Let's call the number of apples you have

[1]

[2]

[3]

Let's solve this ** system of equations**.

[2]

Substituting [2] into [1]:

[4]

Substituting [2] into [3]:

[5]

Substituting [4] into [5]:

Using this result in [2]:

Using this result in [1]:

What we have found is that if you purchase

This solution is the only solution! It was logical to get to it using algebra while randomly guessing the three numbers would have taken some time. Problems can get much more complicated than the one I showed here, so you can begin to understand the importance of solid algebra skills.

Hope this helps to give you an idea of what you will work towards learning when you get into your algebra course. Best of luck!

Featured 3 weeks ago

See below.

**Question:**

What is the vertex form of

There are a couple forms that will fill the requirement.

I believe the form you may want is:

**Form 1)**

where **(h, k)** is the vertex of the parabola.

**a** tells you whether the curve holds water (is cup shaped), if a is positive. And, **a** tells you the curve spills water (is cap shaped), if a is negative. It also tells you how the curve relates to the graph of

Rearrange things a little to compete the square.

Factor.

Now, put the result into the form 1):

The vertex, then, is

There are other useful forms that yield other kinds of information easily and they are used when it is desirable to be able to read that information easily from the equation.

Featured 1 week ago

To solve this, we expand/distribute using FOIL:

As you can see in this image, the first thing we do is multiply the

Then the

Then the

and finally the

Now combine them all together:

We can still simplify the like terms. Let's color-code them:

Combine like terms and get the final answer:

Hope this helps!

Featured 1 week ago

The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve "x^2-4=0rArr(x-2)(x+2)=0#

#rArrx=-2" and "x=2" are the asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),g(x)toc" ( a constant)"# Divide terms on numerator/denominator by the highest power of x

#"that is "x^2#

#g(x)=((3x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)#

#"as "xto+-oo,g(x)to(3+0-0)/(1-0)#

#rArry=3" is the asymptote"#

graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}