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(12w^4- 6w³+w²)/3w² can be rewritten as
12w^4/3x^2 - 6w^3/3w^2 + w^2/3w^2

So just simplify the fractions.
12w^4/3x^2 - 6w^3/3w^2 + w^2/3w^2
=4w^2-2w+1/3

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18r^3-39r^2+20r
Factor out an r
r(18r^2-39r+20)
=r(3r-4)(6r-5)

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Using the rule loga+logb=logab

log(x+1)+log(x-1)=log3
log[(x+1)*(x-1)]=log3
log(x^2-x+x-1)=log3
log(x^2-1)=log3
10^[log(x^2-1)]=10^log3
x^2-1=3
x^2=4
x=sqrt4
x=+2,-2

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x=Japan
y=Switzerland

3x+2y=78.86
2x+3y=74.84
Find x and y

Multiply each equation to cancel one of the variables out.
3x+2y=78.86 x2
6x+4y=157.72

2x+3y=74.84 x3
6x+9y=224.52

Now the two equations have 6x, and if you subtract one from the other, you can cancel out one of the variables.
6x+4y=157.72
6x+9y=224.52

-5y=-66.8
y=-66.8/-5
y=13.36

Now you have y, substitute y into an equation to find x.
2x+3y=74.84
2x+3(13.36)=74.84
2x+40.08=74.84
2x=34.76
x=34.76/2
x=17.38

So tickets in Japan cost $17.38 and tickets in Switzerland cost $13.36

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If 11 coins are tossed what is the probability of getting EXACTLY 2 heads?
Use P(x=k)=nCk*p^k*q^(n-k)
Where p=probability of success and q=1-p, probability of failure

So
P(x=2)=(11C2)(1/2)^2(1/2)^(11-2)
=(11C2)(1/4)(1/512)
=(11C2)(1/2048)
=0.0269

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There are only 4 queens in a deck of cards.
P(a)= probability that event A occurs
= (total number of desired outcomes)/(total number of possible outcomes)
Use combinatorics.

The total number of possible outcomes in a) and b) is
52C5 = 2598960

The total number of desired outcomes in a) is
48C5 = 1712304
It's 48 because you don't want any queens, and there are 4 queens in a deck of 52

The total number of desired outcomes in b) is
0 because there are only 4 queens in a deck of 52, but you want your 5 card hand to be all queens.

So
a)P(no queens in a 5 card hand)=1712304/2598960=0.659
b)P(all queens in a 5 card hand)=0/2598960=0

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Your points are (3,5) and (1,8)
Use the slope formula: m=(y2-y1)/(x2-x1)

m=(8-5)/(1-3)
m=3/-2
m=-3/2

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Probability is (total number of desired outcomes)/(total number of possibly outcomes)

Desired outcome: you want 1 king, and 4 other cards

To get that, you use combinatorics:
4C1 x 48C4 = 778320

Total outcome: 5 card hand

To get that, you do 52C5 = 2598960

Then the probability is
P(1 king in a 5 card hand)= 778320 / 2598960 = 0.299

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2x-7y>14
Determine if the graph is solid or dotted, and shaded above or below.

First off, the graph is dotted. You can tell this because the inequality sign is greater/less than. If the sign was greater/less than or equal to, then the graph would be a solid.

To determine whether the graph is shaded above or below, you must first solve for y, and graph it.





So the equation is
To graph this, look at it with an equal sign instead.
This has a y-intercept of -2, and a slope of 2/7.

Now to determine what part is shaded, look at the equation that is solved for y, and pay attention to the inequality sign. If it is y>... then it is shaded above. If it is y<... then it is shaded below. In this case, it is shaded below.

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Yes that's right, anything to the power of 0 is 1.

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take the log of both sides and then separate the 2(3^x) by the log(m)+log(n)=log(mn) rule.


log(2(3^x))=log(7(5^x))
log(2)+log(3^x)=log(7)+log(5^x)
log(2)+xlog(3)=log(7)+xlog(5)
xlog(3)-xlog(5)=log(7)-log(2)
x(log(3)-log(5))=log(7)-log(2)
x=(log(7)-log(2))/(log(3)-log(5))
x=-2.452427 <--i used a calculator

x=(log(7)-log(2))/(log(3)-log(5)) can be rewritten using the rule log(m)-log(n)=log(m/n)
x=log(7/2)/log(3/5)

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2x–5y=-1
x–3y=-3

You first want to get rid of one of the variables.
There are two different ways of doing this.
The first way is to solve the 2nd equation for x, and then plug it into the 1st equation to solve for y.
The second way is to multiply the 2nd equation by 2, and then subtract this new equation from the 1st equation so that x cancels out. Left with y, you solve for it.

The first way.
2x–5y=-1
x–3y=-3

Solve the 2nd equation for x
x-3y=-3
x=3y-3

Substitute x into the first equation
2x-5y=-1
2(3y-3)-5y=-1

Simplify and solve for y
2(3y-3)-5y=-1
6y-6-5y=-1
y=5

Plug y back into an equation with x in it.
x=3y-3
x=3(5)-3
x=15-3
x=12

The second way.
2x–5y=-1
x–3y=-3

Multiply the 2nd equation by 2
x–3y=-3
(x-3y=-3)x2
2x-6y=-6

Subtract the new equation from the first equation.
2x–5y=-1
2x-6y=-6
--------------
y=5

Plug y into an equation with x in it.
x–3y=-3
x-3(5)=-3
x-15=-3
x=12

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The first step is to multiply the 2nd equation by 2.
This is because in order to solve it, you must cancel out one of the variables.
By multiplying the 2nd equation by 2, you can subtract the new equation (from multiplying the 2nd equation by 2) from the 1st equation, and get 0x. This leaves you with only y. You can then solve for y, and then plug y back into an equation with x and find x.




Multiply the 2nd equation by 2
()x2


Subtract this from the first equation


-------


Solve for y



Plug y back into an equation with x

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Let:
x=James's tickets
y=Ashley's tickets
z=Amanda's tickets

Write your equations
x+y+z=340
x+40=y
y+80=z

Now solve your system...
x+40=y
x-y=-40

x+y+z= 340
x-y =-40
2y+z= 380
z= 380-2y

y+80=z
y+80=380-2y
y+2y=380-80
3y=300
y=100

z=380-2y
z=380-2(100)
z=380-200
z=180

x+40=y
x+40=100
x=60

Check:
x+y+z=340
60+100+180=340

James has 60 tickets.
Ashley has 100 tickets.
Amanda has 180 tickets.

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(x)+(x+2)+(x+4)+(x+6)+(x+8)=-210
x+x+2+x+4+x+6+x+8=-210
5x+20=-210
5x=-230
x=-46

Check:
(-46)+(-44)+(-42)+(-40)+(-38)
=-46-44-42-40-38
=-90-120
=-210

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Solve .

First, take the log of both sides


Then move the exponent to the front of the log (it's a rule).


Now isolate the x.




That's the answer, unless you're allowed a calculator. If so, punch that into your calculator and you should get x=0.357

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d=2k
d=c
c=2k
k=d/2
k=c/2

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The perimeter of a rectangle is 48 cm. The length is 6 cm less than twice the width. Which system would model this situation, and what are the length and the width?

Let x=length and y=width



Solve the 2nd equation for x or y (whichever is more convenient) and then plug into the 1st equation.









Substitute y into a previous equation to solve for x.





The length is 14cm and the width is 10cm.

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The perimeter of a rectangular rose garden is 400 M. The length is 3 M more than twice the width. Find the length and width.

Let x=length, y=width



Solve the 2nd equation for x or y, and then substitute it into the first equation.










Now substitute y into a previous equation and solve for x.



or depending how you want your answer

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Make it so that x is not a fraction

=
=

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The length of a rectangle is 4 inches longer than twice its width. The area of the rectangle is 96 square inches. What is the length and width of the rectangle?

Let x=length and y=width



Solve the first equation for y and substitute it into the 2nd equation










Reject x=-8 because the length cannot be a negative number
So x=6
Substitute x to find y

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Factor out the

=

If it says to solve, equate it to 0 and solve for x



and

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You want to divide 4 7/8 by 1/2

First make the 4 7/8 into an improper fraction
(4*8+7)/8=39/8

Now you have (39/8) divide by (1/2)
Flip the fraction after the divide sign and change the sign into a multiply sign

(39/8) x (2/1)

Now multiply them together
(39/8) x (2/1)
=(39/4)

If it says to make it into a proper fraction again, do so
39/4= 9 3/4

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the perimeter of a college athletic field is 92 meters and the length is 12 m more than the width. find the length and width.
Let x=length and y=width
2x+2y=92
x-12=y

Substitute the 2nd equation into the first one
2x+2y=92
2x+2(x-12)=92
2x+2x-24=92
4x=116
x=116/4
x=29

Substitute x into one of the original equations
x-12=y
29-12=y
17=y

the length is 29m, the width is 17m

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The sum of the ages of a man and his wife is 80 years. 10 years ago, the ages of the wife was exactly two-third the age of the man. Find the present ages of the man and his wife.

Let:
man's age = x
wife's age = y

x+y=80 <--sum of their age
y-10=2/3x <--10 years ago, the age of the woman was 2/3 of the man's current age

y=2/3x+10 <--solve the 2nd equation for y
x+2/3x+10=80 <--plug above into 1st equation
5/3x=70 <--simplify and solve
x=42 <--plug back into original equation to find y
42+y=80
y=80-42
y=38

Since x=42 and y=38
The man's age is 42 and the wife's age is 38

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From a group of eight boys amd two girls a boy and a girl will be selected to attend a conference.
In how many ways can the selection be made?

First choose your boy:
8C1

Then choose your girl:
2C1

Now multiply them together
8C1 x 2C1
So now you get 8x2 which = 16

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Eva is 2 years older than Julienne. The sum of their ages is 20. Find their present ages.

Let:
Eva's age = x
Julienne's age = y

x+y=20 <--the sum of their ages is 20
x-2=y <--Eva is 2 years older than Julienne

You can solve this system by either adding/subtracting or substituting.

Adding/Subtracting:
x+y=20
x-2=y

Rearrange the 2nd equation
x-2=y becomes x-y=2
So now we have

x+y=20
x-y=2
Now add the equations, you will get
2x=22
Now solve...
2x=22
x=11

Now that you have x, plug it into one of the equations to solve for y
x-y=2
11-y=2
11-2=y
9=y

Since x=11 and y=9
Eva's age is 11 and Julienne's age is 9
Substituting:
x+y=20
x-2=y

Since the 2nd equation is already arranged in a way that a variable = something... substitute it into the 1st equation.
x+y=20
x+(x-2)=20
x+x-2=20
2x-2=20
2x=22
x=11

Now that you have 11, plug x into your 2nd equation
x-2=y
11-2=y
9=y

Since x=11 and y=9
Eva's age is 11 and Julienne's age is 9

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Solve the system



Solve the 2nd equation for y



So now you have



Now substitute the 2nd equation into the 1st one


Now expand and group like terms




Now factor and solve




 
    

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f(-9)=6
f(7)=10
f(x-9)=

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x=-7, 1
0=(x+7)(x-1)

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The points that are given to you are (5,-5) and (0,8).
The equation is y=mx+b.
What you need to find is the slope, m and the y-intercept, b.

To find the slope, m, you use the formula:

So,



Now from y=mx+b, we have
To find the y-intercept, we can substitute one of the points in for x and y, and solve for b.

(0,8)





Now from y=mx+b, we have