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2

Answer:

#a=12#

Explanation:

#color(blue)("Important fact")#

A fraction's structure is such that we have:

#("numerators")/("denominators") color(white)("d")->color(white)("dd") ("count")/("size indicator of what is being counted")#

You can not #color(purple)(ul("DIRECTLY"))# add, subtract or compare the 'counts' unless the 'size indicators' are the same.

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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

#color(green)(a/15=[4/5color(red)(xx1)] color(white)("dddd")->color(white)("dddd")a/15=[4/5color(red)(xx3/3) ])#

#color(green)(color(white)("dddddddddddddddd")->color(white)("dddd")a/15=12/15)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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: #a=12#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Check")#

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

#12/15-:4/5#

#12/15xx5/4 #

# 5/15xx12/4 #

#1/3xx3/1=1#

3

Answer:

#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)#

Explanation:

Given:

#x^3+y^3#

Note that if #y = -x# then this is zero. So we can deduce that #(x+y)# is a factor and separate it out:

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

We can calculate the discriminant for the remaining homogeneous quadratic in #x# and #y# just like we would for a quadratic in a single variable:

#x^2-xy+y^2#

is in standard form:

#ax^2+bxy+cy^2#

with #a=1#, #b=-1# and #c=1#

This has discriminant #Delta# given by the formula:

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

Since #Delta < 0#, this quadratic has no linear factors with real coefficients.

We can factor it with complex coefficients by completing the square and using #i^2=-1# as follows:

#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)#

1

Answer:

See a solution process below:

Explanation:

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

He adds 3 to the number #=> n + 3#

Then doubles the result #=> 2(n + 3)#

Finally, he subtracts 7 #=> 2(n + 3) - 7#

He ended with 9 #=> 2(n + 3) - 7 = 9#

We can solve this equation by first adding #color(red)(7)# to each side of the equation to isolate the term with parenthesis while keeping the equation balanced:

#2(n + 3) - 7 + color(red)(7) = 9 + color(red)(7)#

#2(n + 3) - 0 = 16#

#2(n + 3) = 16#

Next, divide each side of the equation by #color(red)(2)# to eliminate the need for parenthesis while keeping the equation balanced:

#(2(n + 3))/color(red)(2) = 16/color(red)(2)#

#(color(red)(cancel(color(black)(2)))(n + 3))/cancel(color(red)(2)) = 8#

#n + 3 = 8#

Now, subtract #color(red)(3)# from each side of the equation to solve for #n# while keeping the equation balanced:

#n + 3 - color(red)(3) = 8 - color(red)(3)#

#n + 0 = 5#

#n = 5#

Jaden started with the number: #color(red)(5)#

He adds 3 to the number #=> color(red)(5) + 3 = 8#

Then doubles the result #=> 2 xx color(red)(8) = 16#

Finally, he subtracts 7 #=> color(red)(16) - 7 = 9#

4

Answer:

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

Explanation:

  • 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...

1

Answer:

# color(blue)( C_5^7 = (7!)/(5! * (7-5)!) #
# color(brown)( C_5^7 = 21 #

Explanation:

Questions:
How do you find the value of the combination
# color(blue)( C(7,5) = C_5^7 # ?

# color(blue)( C_5^7 # can be read in English as, "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.

# color(blue)( C_5^7 = (7!)/(5! * (7-5)!) #
# C_5^7 = (7*cancel(6)3*cancel(5*4*3*2*1))/(cancel(5*4*3*2*1) * (cancel(2)*1)) #

# color(brown)( C_5^7 = 21 #

1

Answer:

2 real solutions

Explanation:

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

Quadratic equation form: #ax^2+bx+c#, in this case #a# is 2, #b# is 1 and #c# is -1

Discriminant: #b^2-4ac#

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

#1^2-4*2*-1#

#1-4*2*-1#

#1-(-8)#

#9 rarr# A positive discriminant indicates that there are 2 real solutions (the solutions can be positive, negative, irrational, or rational, so long as they are real)

Negative discriminants indicate that the quadratic function has 2 imaginary (involving #i#, the square root of -1) solutions.

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 #(x+6)^2#, which has a discriminant of 0)

3

Answer:

Letters!

Explanation:

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 #2# less apples than the first sibling.

[3] The total number of apples you want to buy at the store is #10#.

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 #a#, the number of apples your first sibling has #b#, and the number of apples your second sibling has #c#. We can write the equations corresponding to each condition as:

[1] #=> 1/2a = b#

[2] #=> c = b - 2#

[3] #=> a+b+c = 10#

Let's solve this system of equations.

[2] #=> b = c+2#

Substituting [2] into [1]:

[4] #=> a = 2c+4#

Substituting [2] into [3]:

[5] #=> a+2c=8#

Substituting [4] into [5]:

#=> 2c+4+2c=8#

#=> 4c + 4 = 8#

#=> 4c = 4#

#=> color(blue)( c = 1)#

Using this result in [2]:

#=>1 = b - 2#

#=> color(blue)(b = 3)#

Using this result in [1]:

#=>1/2a = 3#

#=> color(blue)(a = 6)#

What we have found is that if you purchase #10# total apples, you would keep #6# apples, your first sibling would get #3#, and the second sibling would get #1#.

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!

1

Answer:

See below.

Explanation:

Question:
What is the vertex form of
# color(blue)( y = x^2 + 14x + 3 # ?

There are a couple forms that will fill the requirement.

I believe the form you may want is:
Form 1) # color(blue)( (y - k) = a(x - h)^2 #,
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 # y = x^2 #, whether fatter or skinnier.

# color(blue)( y = x^2 + 14x + 3 #
Rearrange things a little to compete the square.
# y - 3 = x^2 + 14x #
# y - 3 + 49 = x^2 + 14x + 49 #
Factor.
# (y + 46 ) = (x + 7)^2 #
Now, put the result into the form 1):
# color(purple)( (y - (-46)) = 1 * (x - (-7))^2 #

The vertex, then, is # color(purple)( (-7, -46) #

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.

1

Answer:

#a^2 - a - 30#

Explanation:

To solve this, we expand/distribute using FOIL:
enter image source here

As you can see in this image, the first thing we do is multiply the #color(steelblue)"firsts"#:
#a * a = a^2#

Then the #color(purple)"outers"#:
#a * -6 = -6a#

Then the #color(peru)"inners"#:
#5 * a = 5a#

and finally the #color(olivedrab)"lasts"#:
#5 * -6 = -30#

Now combine them all together:
#a^2 - 6a + 5a - 30#

We can still simplify the like terms. Let's color-code them:
#a^2 quadcolor(red)(-quad6a quad+quad 5a) quad - quad30#

Combine like terms and get the final answer:
#a^2 - a - 30#

Hope this helps!

1

Answer:

#"vertical asymptotes at "x=+-2#
#"horizontal asymptote at "y=3#

Explanation:

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]}