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A and B are endpoints of the diameter

The radius of the circle =1

So AB  = 2

Note that [BEFA] = [CEFD]

The line is y= 2x 

Any line of the form y = ax will split the rectangle ( a square) into = areas and will have a y intercept  = 0

Triangle HQN sinilar to triangle FQG           Triangle MPF similar to triangle GPH

HQ/HN = FQ/FG                                                MF/PF = GH/PH                                       

HQ/2 = FQ/4                                                       2/PF = 4/PH

HQ/FQ = 2/4 = 1/2                                              PF/PH = 2/4 = 1/2

HQ = ( 1/3)FH

PF = (1/3)FH

FH - HQ - PF  = PQ

FH - (1/3)FH - (1/3)FH = (1/3)FH = PQ

PQ / FH =  (1/3)FH / FH  = 1/3  

Using polynomial interpolation, we can find that f(x) = -17/3*x^3 + 59/2*x^2 - 209/6*x + 5.

Therefore, the sum of the coefficients is -17/3 + 59/2 - 209/6 = -11.

Hope this helps!

In a row of five squares, each square is to be colored either red, yellow, or blue, so that no two consecutive squares have the same color. How many ways are there to color the five squares, if there must be at least three yellow squares?    

In order not to have two yellows next to each other,    

the three yellows must occupy squares 1, 3, and 5.    

That leaves squares 2 and 4 to fill, and two colors    

to do it with. There are only four ways.   

Blue Blue     Red Red     Blue Red     Red Blue    

So, the answer is four ways.     

_.    

\(\cfrac{1}{1 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{2}}}} \)

Work from bottom  to  top

4 + 1/2 = 9/2

1/ (9/2)  = 2/9

4 + 2/9  = 38/9

1/(38/9) = 9/38

1 + 9/38  = 47/38

1/(47/38) = 38 / 47

A baking club wants to form an executive committee. There are $20$ people in the baking club, including Mark. In how many ways can the baking club form an executive committee with $7$ people, including Mark?     

₂₀C₇ = 77,520    

This is called the combination of 20 items taken 7 at a time.    

Sometimes called the shortened form of "20 choose 7" but    

I prefer the first way.   

_.    

Let a1, a2, a3, ..., a10 be an arithmetic series. If a1 + a3 = -2 and a2 + a4 + a6 = 1, then find a1.    

call "d" the amount of addition between successive terms.    

starting with a₁   then  a₂  =  a₁ + d                                              =   a₁ + d     

                                    a₃  =  a₂ + d  =  (a₁ + d) + d                     =   a₁ + 2d    

                                    a₄  =  a₃ + d  =  (a₁ + d + d) + d               =   a₁ + 3d    

                                    a₅  =  a₄ + d  =  (a₁ + d + d + d) + d         =   a₁ + 4d    

                                    a₆  =  a₅ + d  =  (a₁ + d + d + d + d) + d   =   a₁ + 5d    

given       a₁ + a₃  =  – 2    

                a₁ + a₁ + 2d  =  – 2    

                2a₁ + 2d  =  – 2    

                     a₁ + d  =  – 1                                   (eq 1)    

given        a₂ + a₄ + a₆  =  1    

                (a₁ + d) + (a₁ + 3d) + (a₁ + 5d)  =  1    

                3a₁ + 9d  =  1                                       (eq 2)    

multiply (eq 1) by – 9         – 9a₁ – 9d  =  9    

add (eq 2)                             3a₁ + 9d  =  1    

                                           – 6a₁          =  10    

                                                a₁          =  – 10 / 6    

                                                        a₁  =  – 5 / 3    

_.   

Every item in the first term is multiplied with every item in the second term.    

Pick out the items that, when multiplied, result in an "x" coefficient product.    

(x3 + x2 + x + 1) • (x4 – 8x3 + 17x2 – 23x + 14)     gives you + 14x    

(x3 + x2 + x + 1) • (x4 – 8x3 + 17x2 – 23x + 14)     gives you – 23x    

add them together for the final value                                     – 9x    

_.    

Let's multiply the polynomials and see what we get. 

We have \((x^3)(x^2 + tx)\)

Distributing in the x^3, we find that we get the expression \(x^5+tx^4\)

there is actually no x^2 term, meaning t can pretty much be anything. 

As long as t doesn't contain a x^-2 term, then it isn't possible there is an x^2 term. 

Thanks! :)

A line and a circle intersect at A and B as shown below. Find the distance between A and B.    
The line is x = 4, and the equation of the circle is x^2 + y^2 = 25.    

Substitute 4 for x in the equation of the circle to find the point(s) common to both curves.    

x² + y²  =  25    

       y²  =  25 – x²    

       y²  =  25 – 4²    

       y²  =  9    

       y  =  + 3       —>>  The curves intersect at (+4, +3) and (+4, –3)    

Since x=4 is a straight, vertical line, there is no horizontal component of the distance between A and B.    

Therefore, the distance between + 3 and – 3 is 6 units.    

_.    

Every item in the first term is multiplied with every item in the second term. 

Pick out the items that, when multiplied, result in an "x" coefficient product.  

(x³ + x² + x + 1) • (x⁴ – 8x³ + 17x² – 23x + 14)     gives you + 14x    

(x³ + x² + x + 1) • (x⁴ – 8x³ + 17x² – 23x + 14)     gives you – 23x    

add them together for the final value                                     – 9x    

_.    

For this problem, we have to go throug every function and find when f(x) = x. 

First off, we have \(f(x)=2x+8\). 

We have to solve \(2x+8=x\), in which we get \(x=-8\). However, -8 isn't in the range \(1 \le x \le 2\), so it isn't a valid solution. 

Next, we have \(f(x) = 13 - 5x\)

Solving for \(13-5x=x\), we get \(x=13/6\). Indeed, 13/6 is in the range \(2 < x \le 3\), so it is a valid solution. 

After, we have \(f(x) = 20 - 14x\)

Solving for \(20-14x=x\), we get \(x=4/3\), which is not a valid solution since it's not in the range \(3 < x \le 4\). 

Lastly, we have \(f(x)= 40 + 5x\). 

We have \(40+5x=x\), in which we get \(x=-10\), which is not a valid solution. 

So, we only have x = 13/6

Thanks! :)

For there to be no real solutions, the discriminant needs to be NEGATIVE

The descriminant is \(b^2 - 4ac <0 \)  with  \(b = k ;   \space a = 5   ; \space c = 8\)

Plugging it all in, we have

\(k^2 - 4(5)(8) <0 \\ k^2 < 160\)

\(k <   12.65\)      

We are trying to find the largest INTEGER, which would be 12. 

Thanks! :)

\(XZW \) will form a triangle \(\triangle XZW \)

\(\angle ZXW  = 90^\circ \\ \angle XWZ = 60^\circ\)

So \(\angle XZW =  180^\circ - 90^\circ - 60^\circ  =   30^\circ\)

Thanks! :)

Let G be Grogg's current age, let D be Grogg's dad's current age.

1. "Seven years ago, Grogg's dad was 9 times as old as Grogg":
   D - 7 = 9(G - 7). We can rearrange and rewrite to get:
   D = 9G - 56.
 

2. "Four years ago, Grogg's dad was 6 times as old as Grogg":
   D - 4 = 6(G - 4), We can rearrange and rewrite to get:
   D = 6G - 20.

Now we have the two equations:
1. D = 9G - 56
2. D = 6G - 20

Next, we can set the right-hand sides of Equation 1 and Equation 2 equal to each other, since we know their left hands sides are equal:
9G - 56 = 6G - 20
9G - 6G - 56 = -20,
3G = 36, 

So we have: G = 12
 

Now that we have Grogg's age, we can find his dad's age. Using the first equation:
D = 9G - 56 = 9(12) - 56
D = 52
 

Thus, Grogg's dad is currently 52 years old. 

You can check this answer with equation 2, if you'd like. 

Having a units digit of 2 and a remainder of 1 means the units digit of the quotient is 1. So the problem is asking "How many 3 digit integers have a units digit of 1 and are divisible by a number 2-9?"

Because 2, 4, 5, 6, 8 dont have any multiples that end in 1, we can exclude those

Because 9 is a multiple of 3, counting the 3's already includes 9's so we can exclude 9 too

To find the number of 3 digit integers that end in 1 and are divisible by 3 and/or 7, we can see that every 3 terms in the sequence 101, 111, 121, 131, ... is divisible by 3, so that is simply 90 (The number of terms in the sequence)/3 = 30.

To find the number of 3 digit integers that are divisible by 7, we can find that, extending the sequence to 1001, there are 91/7 = 13 terms, and subtracting the 1001 gives 12.

This logic counts integers that are divisible by 21 twice, so subtracting ⌊90/21⌋ = 4 would even it out.

Finally, the sum turns out to be 

30(Multiples of 3) + 12 (Multiples of 7) - 4 (Multiples of 21) = 38 3 digit numbers