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If l1 is horizontal to (1,1), then the equation of the line is y = 1.

If l2 is horizontal to (2,2) then the equation of the line is x = 2.

the intersection must be 2,1.

i hope this helped.

gavin

Did you understand any of it?

It would have been good if you had told me more specifically what you did not understand.

It is a 4 digit number but 1 digit (exactly one digit) is repeated just once.

So 3 digits out of 10 have to be chosen.

There are 10C3 = 120 ways to do this.

say the digits chose are  5,6 and 9, they can be in any order and there are 3! possible orders =6 orders.

So so far that is   120*6=720 ways to get a 3 digit number where the digits are all different.

But we want a 4 digit number.

One digit has to be repeated there are 3 possibilities so that is   720*3 = 2160 posibilities.

But there are 4 postions that extra number can be put in. So that is  2160*4=8640 numbers

But I have double counted because    1434 is the same if I swap the 4s around so I have to halve this answer

8640/2 = 4320    

So the number of favourable outcomes is 4320.

Altogether there are  10*10*10*10 = 10000 4 digit numbers (if they are allowed to start with 0)

so the prob of getting just one digit repeated once is   4320/10000 = 0.4320 = 43.2%

Do you understand now?

You will need to write your query on this thread but also send me a private message with the address of this thread attached. Otherwise I will not see it. 

If you do have a query try to ask a specific question.  

Anyway I hope that you understand :)

The only way I can think to do this is with co-ordinate geometry.

Equation of AF is    y=(-4/3)x+4

Equation of EB is    y=(-2/7)x+2

Simultaneous solution of these gives the point  D which is \((\frac{21}{11},\frac{16}{11})\)

Distance AB can be found using Pythagoras's theoem.  It is  \(\sqrt{65}\)

Equation of AB is 

  \(y=\frac{-4x}{7}+4\\ 7y=-4x+28\\ 4x+7y-28=0 \)

Now I can use the perpendicular distance formula of a point to a line to find the perpendicular height between AB and D

\(d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\ d=\frac{|4*\frac{21}{11}+7*\frac{16}{11}-28|}{\sqrt{16+49}}\\ d=\frac{|4*\frac{21}{11}+7*\frac{16}{11}-28|}{\sqrt{16+49}}\\ d=\frac{112}{11}*\frac{1}{\sqrt{65}}\\ d=\frac{112\sqrt{65}}{11*65}\\ d=\frac{112\sqrt{65}}{715}\\ \)

Area of ABD = 

\(\frac{1}{2}*\sqrt{65}*\frac{112\sqrt{65}}{715}\\ =\frac{56*65}{715}\\ =\frac{56*13}{143}\\ =\frac{728}{143}\\ =5\frac{13}{143}u^2\)

That is assuming I have made not stupid errors.

You could check this using heron's formula but that would be a lot of work too. 

Since the area of ABC is 14 my answer does sound reasonable. 

In a class of 28 students, the teacher selects four people at random to participate in a geography contest. What is the probability that this group of four students includes at least two of the top three geography students in the class? Express your answer as a common fraction.

I think the number of ways that choice can be made is ...

\(\binom{3}{2}*\binom{25}{2}+\binom{3}{3}*\binom{25}{1}\\ =3*300+1*25\\ =900+25\\ =925\)

And the total number of possibilities is  28C4 = 20475

So the prob is    \(\frac{925}{20475} = \frac{37}{819} \)

The answer in simplest radical form.

Each of eight houses on a street is painted brown, yellow or white. Each house is painted only one color and each color is used on at least one house. No two colors are used to paint the same number of houses. In how many ways could the eight houses on the street be painted?

Well this is what I am thinking

Let the colours be A B and C     

    There are 3!=6 possible variations on which colour goes with which letter

Now the house numbers can be  5,2 and 1    OR   4,3,1

    There can be no other combination.

so we have

\(6(\frac{8!}{5!2!}+\frac{8!}{4!3!})=6(168+280)\)

=2688 possible combination.

Please, if anyone can help and give a good explanation to back up their answer, I would greatly appreciate the help. Thank you!

Ok! thanks for the detailed explanation

Thank you guys! GYanggg, even though you got it wrong, good job, I really appreciate your and guest’s help.

First of all, your second equation is the exponential one but is written incorrectly. It should look like this:

an = 35(1.10)^(n − 1). So, in 50 years you will have:

a(50) = 35(1.10)^(50 - 1)

a(50) = 35(1.10)^49

a(50) = 35 x 106.7189571......

a(50) = $3,735.16 - The value of stock B in 50 years.

an = 10n + 152

a(50) =10 x 50 + 152

a(50) =500 + 152

a(50) =$652 - This is the value of stock A in 50 years

Stock B - stock A =$3,735.16 - $652 =$3,083.16 - This is how much more will stock B worth than stock A in 50 years. Or, "D" on your list of multiple choices.

Solution:

the volume of can: 753.6 cm

The diameter of can: 8 cm

So, h=V/pir^2, so plugging the value in, we get 14.99 (rounded). 

What is the volume of a cylinder?

V=πr^2h, where r=radius, h=height.

Calculate the volume of cylinder A and B and compare them.

Gavin: Your solution dosen't balance the equation!!!!

Solve for x:

x^3/27 - 3 x^2 + 81 x - 729 = 64

Subtract 64 from both sides:

x^3/27 - 3 x^2 + 81 x - 793 = 0

The left hand side factors into a product with three terms:

1/27 (x - 39) (x^2 - 42 x + 549) = 0

Multiply both sides by 27:

(x - 39) (x^2 - 42 x + 549) = 0

Split into two equations:

x - 39 = 0 or x^2 - 42 x + 549 = 0

Add 39 to both sides:

x = 39 or x^2 - 42 x + 549 = 0

Subtract 549 from both sides:

x = 39 or x^2 - 42 x = -549

Add 441 to both sides:

x = 39 or x^2 - 42 x + 441 = -108

Write the left hand side as a square:

x = 39 or (x - 21)^2 = -108

Take the square root of both sides:

x = 39 or x - 21 = 6 i sqrt(3) or x - 21 = -6 i sqrt(3)

Add 21 to both sides:

x = 39 or x = 21 + 6 i sqrt(3) or x - 21 = -6 i sqrt(3)

Add 21 to both sides:

x = 39      or      x = 21 + 6 i sqrt(3)      or      x = 21 - 6 i sqrt(3)

I already answered, but here it is:

The cylindrical storage container as a volume of:

\(5^2\pi\cdot7\approx550\)

Since 550 is greater than 500.

To answer your question,

The cat litter will fit, since the volume of the cat liter is less than the storage. 

The cylindrical storage container as a volume of:

\(5^2\pi\cdot7\approx550\)

Since 550 is greater than 500.

To answer your question,

The cat litter will fit, since the volume of the cat liter is less than the storage. 

On the left side of the equation, we can tell that that we can easily factor it. 

It is a perfect cube. 

We use \((x-y)^3=x^3-3x^2y+3xy^2-y^3\)

Then we have:

\((\frac{x}{3}-3)^3=64\)

\(\frac{x}{3}-3=4 \)

x = 21

I hope this helped,

Gavin

1) Which shape best describes the object generated when half circle is rotated about the axis?

*edit again*

D: Solid Sphere

\(\text{____________________________________________________________________}\)

2) Which three-dimensional object is formed when the shape is rotated about the axis as shown?

C: Cylinder  (it is a short cylinder, like a hockey puck)

\(\text{____________________________________________________________________}\)

3) Which description best describes the three-dimensional object that is formed when the shape is rotated about the axis as shown?

*edit*

At first I said A: A flat disk with a hole in the center, but actually I think that you're right...it is  D: A hollow cylinder! I had to look up what a hollow cylinder was!

\(\text{____________________________________________________________________}\)

4) Which figure shows how a shape can be rotated about an axis to form a cylinder?

D:   

\(\text{____________________________________________________________________}\)

5) Which figure shows how a shape can be rotated about an axis to form the given object?

D:

Hi supermanaccz!

So from 100 to 200, we first need to determine what numbers are bad. 

From 100 to 159, they are all good, since we can have 1:00 and 1:59.

But from 160 to 199, they are all bad. There is no 1:60, since that will be 2 o'clock. 

There is no 1: 99.

So we need to calculate how many numbers from 160 - 199 are equal to 3 mod 4.

The smallest number that is from 160 - 199 that is equal to 3 mod 4 is 163.

3 / 4 = 0 R 3

163 / 4 = 40 R 3

The largest is 199. 

199 / 4 = 49 R 3

To calculate how many numbers we want from 163 to 199, we do:

\((199-163)/3+1=\boxed{13}\)

That will be your final answer .

I hope this helped, 

gavin.

First, in this problem, you would assume that there are five numbers to multiply since Dwayne played 5 games. 

In the first blank, he could choose any move, and there are three, rock, paper, and scissors. In the first blank, you would put 3. Then, since you already picked one, you can't repeat it, so you would end up with 2 for the second blank. Continuing with the next blanks, the third, fourth, and fifth spaces would be 2 choices because you cannot have the one choice you chose previously. In the end, the blanks would look somewhat like this:

3    2    2    2    2

And you would multiply all the numbers.

3  x  2  x  2  x  2  x  2

                                                                   = 6 x 8

                                                                   = 48

Can someone help me on this problem?!?

Part A:

Let's condense the information we know. I find this small regrouping step helps me immensely.

• There are 65 adults
• There are 15 children
• 41 out of 65 adults ate popcorn
• 6 out of 15 children ate popcorn
• There are 80 people in total

You can even make a quick table, but I think there is not enough information where this is necessary.

\(P(\text{Child or Bought Popcorn})=P(\text{Child})+P(\text{Bought Popcorn})-P(\text{Child and Bought Popcorn})\\ P(\text{Child or Bought Popcorn})=\frac{15}{80}+\frac{47}{80}-\frac{6}{80}=\frac{56}{80}=\frac{7}{10}=0.7\)

Part B:

\(P(\text{Adult or No Popcorn})=P(\text{Adult})+P(\text{No Popcorn})-P(\text{Adult and No Popcorn})\\ P(\text{Adult or No Popcorn})=\frac{65}{80}+\frac{80-(41+6)}{80}-\frac{65-41}{80}\\ P(\text{Adult or No Popcorn})=\frac{65}{80}+\frac{33}{80}-\frac{24}{80}=\frac{74}{80}=0.925 \)

I hope this helps you! This is just algebraic manipulation.