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Each cube has   3/6  even numbers    or   1/2 chance of being even when rolled

1/2  x 1/2 = 1/4    chance of BOTH being EVEN

For quadratic Formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a = 3  b=-12 and c = -663

You can simplify the math.....and maybe see 'in your mind' if you can factor it by first dividing the left and right sides of the equation by 3 to get

x^2 - 4x - 221 =0

     Doesn't help me much

        use the QF 

            a = 1 b= -4 c= -221  

4+-sqrt(16-4(1)(-221)   /  2   =     4+- 30   /  2   ========>   17    or   -13

SO factoring would look like    (x-17)(x+13) = 0

The smallest possible value of  z=6, so that:

w^3+x^3+y^3 =z^3

3^3 + 4^3 + 5^3 = 6^3

27 + 64 + 125 = 216

Since the tank holds 200 gallons ad it has 140 gallons in it, there are only 200-140 = 60 gallons left to fill.

   60 galllons/ 2gal/min = 30 minutes

\(\text{we normalize the data by finding }\\ z = \dfrac{X - \mu}{\sigma}\\ z_1 = \dfrac{120-180}{30} = -2\\ z_2 = \dfrac{240-180}{30} = 2\\ \text{using the 68, 95, 99.7 rule we have that}\\ P[-2 \leq z \leq 2] = 0.9545 \\ \text{we want the probability outside this area so we have}\\ P([z<-2) \wedge (2 < 2)] = 1-0.9545 = 0.0455\)

\(u = -7 - 2v\\ 3u = 4v - 25 \\ 3(-7-2v)=4v-25\\ -21-6v=4v-25\\ 4=10v \\ v=\dfrac 2 5 \\ u = -7 - 2\left(\dfrac 2 5\right) = -\dfrac{39}{5}\)

\(7𝑥 + 2𝑦 = 11 \)

\(<=> 𝑦 =- \frac{7x}{2} + \frac{11}{2}\)

We will call \(λ=-7/2 \) is the coefficient of x 

so every parallel line to AB will have \(λ=-7/2\)

lines have type : \(y-y1=λ(x-x1) \) We want coordinates (-5, 25/2 ) 

so x1=-5 and y1=(25/2) 

\(y-(\frac{25}{2})=-\frac{7}{2}(x-(-5))\)

\(y=-\frac{7}{2}(x+5))+\frac{25}{2}\)

\(y=-\frac{7}{2}x-\frac{35}{2}+\frac{25}{2}\)

\(y=-\frac{7}{2}x-\frac{10}{2}\)

\(7x+ 2y=-10\)

"Given the line AB passes through the point (k, k+1) find the value of the constant k."

This means  x= k, y=k+1 verify the equation 7𝑥 + 2𝑦 = 11 

So 

\(7k+2(k+1)=11 \)

\(7k+2k+2=11 \)

\(9k=9\)

so \(k=1\) 

Finally the point (k, k+1) is (1,2) 

Help its helps! 

Here ya go:

2pi is a complete circle .....every 2 pi brings you back to the same starting point....

4pi /2  is 2 pi ...a complete circle....

   so you can simplify 11pi/2   as    (4pi+ 4pi + 3pi) /2     then you drop the 4pi's to simplify to 3pi/2

Here is my attempt at this:
5 bags x 16 candies =80 candies.
80 / 4 = 20 candies per child
Since his 4-year old sister gets 20 -12 =8 candies, then:
8/4 =2 candies per year for each child's age.
Since Cameron will gain 20 + 2 =22 candies, then:
22 / 2 =11 years is Cameron's age.
22 candies of Cameron + 8 candies for youngest sister=30
80 - 30 = 50 candies for the remaining older children.
Since Cameron is the 2nd youngest, it means the remaining 2 siblings must be 12 and 13 years respectively:
So that: 12 x 2 = 24 candies for the 3rd oldest
And 13 x 2 = 26 candies for the oldest child. So, the total distributions of all the candies would be:
8 + 22 + 24 + 26 =80 candies.
So, the oldest child must be 13 years old.

No problem :)

the probabilities must sum up to 1

\(1 = P[1]+P[2]+P[3]+P[4]+P[5] = \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6} + k\\ \\ 1 = \dfrac{20+15+12+10}{60}+k\\ \\ 1 = \dfrac{57}{60}+k\\ \\ k=\dfrac{3}{60} = \dfrac{1}{20}\)

\(\because \angle DHE=\angle FHJ \) (vertical angles) 

\(\because EH=HF\) (midpoint)

\(\because \overline{DE}\parallel\overline{FJ} \Rightarrow \angle EDJ =\angle FJD\)  (alternating interior angles)

\(\therefore \triangle HDE\cong\triangle HJF\) (AAS Congruency) 

\(\therefore [HDE]=[HJF] \Rightarrow [CDHF]+[HDE]=[CDHF]+[HJF]\)

\([CDJ]=[CDEF]\)

\(\boxed{36}\)

I hope this helped,

Gavin. 

can u explain how k= 1/20 plz im still very confused, thanks

1. The expression can be re-written as \((1+\frac1{1000})^{1000}\). Since Euler's constant, \(e\), is equilavent to \((1+\frac1n)^n\) when \(n\) approches infinity. Therefore, the expression is approimately \(e=2.717\cdots\). 

2. \(\frac92\) (I'll add the solution if I have time). 

It's pretty straightforward.

P(X) must sum to 1.  You're given the first 4 values.  Determine the 5th.  The rest are 0.

\(E[X] = \sum~x p(x) = \sum \limits_{x=1}^4~x\cdot \dfrac{1}{x+2}+5k\)

Use the value of \(k=P(5)\) and evaluate that sum.

Thank you so much! It really means a lot!

Times 3 would only be 1500 pages.....you need 1512, so you will have to buy a 4th ream of paper, though you only need 12 more pages from it....

Okay. I mean I guess I won't need them if we are just plugging the numbers in.

We have the equation \(40c + 12a \leq 500\).

Plugging in \(4\) for c and \(2\) for a,

we get:

\(40(4) + 12(2) \leq 500\)

\(= 160 + 24 \leq 500\)

\(= 184 \leq 500\).

Since \(184\) is less than \(500\), the boats are safe. 

Therefore, Judah's family can ride the bumper boats safely.

So the answer is \(\boxed{yes}\).

40(4)+ 12(2)  <= 500?

160 + 24 = 184     Yes it  is safe to ride  

....agree with guest's answer with one small issue:  the volumes should be CUBIC inches (not square inches)

Can I ask what values 40, 12, and 500 stand for?

Isn't it times 3?

12 copies x 126 pages = 1512 pages

will require 4 reams of paper     $ 3.97 x 4 =$15.88    (you will have 488 pieces of paper left over)

what a bizarre question...

Well let's get the equation of the body in standard form so we can see what we're dealing with.

\(x^2+y^2 - 10x - 12y + 45 = 0\\ \text{complete the two squares on the left hand side}\\ (x-5)^2 - 25 + (y-6)^2 - 36 + 45 = 0 \\ \\ (x-5)^2 + (y-6)^2 = 16 \\ \\ \text{so the body is a circle centered at }(5,6) \text{ with a radius of 4 cm}\\ \\ \text{The overall height is 14cm, 8cm of this is the body, so the diameter of the head is 6cm}\\ \text{And the head must be centered at }(5, 11), \text{ i.e. one radius above the top of the body}\\ \\ \text{so the equation of the head centered at }(5,11) \text{ with radius 3 is}\\ (x-5)^2 + (y-11)^2 = 9\)