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l b l   + b^3         b = -2

l - 2 l   +  (-2)^3    =

2  +   ( - 8)    =

-6

13^n    -  6^(n - 2)

Show it's true for n = 2

13^2 - 6^(2 - 2)   =

169 - 1  =  168

168 / 7   =  24

Assume that this is true for  n = k     where k ≥ 2

That is

13^k - 6^(k - 2)      is divisible by 7

Prove it's true for k + 1

That is

13^(k + 1) -  6^ (k + 1 - 2)   is divisible by 7       

[ note ....6^(k + 1 - 2)  = 6^(k - 2 + 1) ]

So we have 

13^(k+ 1)  - 6 ^( k - 2 + 1)

13^k * 13^1  -  6^(k-2) * 6^1

13 * 13^k -   6 * 6^(k - 2)

(6 + 7) 13^k - 6 * 6^(k - 2)

6 [ 13^k - 6^(k - 2) ]   + 7 * 13^k

And since we assumed that 13^k - 6^(k - 2) was divisible by 7, then the first term is divisible by 7, as well

And  the second term, 7*13^k,    is divisible by 7

g(x) is just f(x) shifted up 3 units

So

g(x)  =   f(x) + 3

Since the area of the trapezoid   ZWCD is 120

The height of this trapezoid is AB and the bases are ZD and WC

We have

120 = (1/2) AB  (ZD + WC)

120 = (1/2) 12 ( ZD + 6)     multiply through by 2

240 = 12 (ZD + 6)      divide both sides by 12

20 =  ZD + 6       ubtract 6 from both sides

14 = ZD

Now   angle DBW = angle BDZ

And angle ZQD = angle BQW

And since AZ = WC

Then  ZD = BW

So by AAS....triangle ZQD is congruent to triangle WQZ

And the altitude of trangle DQZ    = altitude of triangle BQW

And since twice these altitudes = AB....then each altitude = 1/2(AB) = 6

So....the area of triangle BQW   =

(1/2) BW * altitude of BQW   = 

(1/2) (14)(6) =

(1/2) (84) =

42 units^2

You can draw triangle \(PAS\) with \(DS\) perpendicular(the altitude) to \(AP\). You know \(PS = 10\). Let's name the midpoint of \(PS\) the letter \(F\). \(PF = 5, PA = 13, \) so \(AF = 12\). Now, cos \(PAS = \frac{AD}{AS}\). We already know \(AS = 13,\) so we just need to solve for \(AD\). To find \(AD\), we first need to solve for \(DS\), which is one of the altitudes of triangle \(PAS\). We know that the area of a triangle is \(\frac{1}{2}bh\), so we can make an equation. \(\frac{1}{2}\times10\times12=\frac{1}{2}\times13\times DS\). From this, we get \(DS = \frac{120}{13}\). Now, we can use Pythagorean Theorum to find \(AD\). \(12^2-(\frac{120}{13})^2=AD^2\). After calculating this, we can get \(AD = \frac{12\sqrt{69}}{13}\). Now, lets plug this into our fraction. We can get \(\frac{\frac{12\sqrt{69}}{13}}{13}\) which simplifies to \(\frac{12\sqrt{69}}{169}\).

- Daisy

never mind. i solved it by myself.

Hi Guest, 

You should not delete your question, because other people may have that question as well.

- Daisy

Simplify the following:

((sqrt(sqrt(24) + 7))/2 + (sqrt(7 - sqrt(24)))/2)^2

[1/2 (1 + sqrt(6)) + 1/2 (sqrt(6) - 1]^2

(sqrt(6) + 1)/2 + (sqrt(6) - 1)/2

(sqrt(6) + 1)/2 + (sqrt(6) - 1)/2 = ((sqrt(6) + 1) + (sqrt(6) - 1))/2:

[(1 + sqrt(6) - 1 + sqrt(6))/2]^2

1 + sqrt(6) - 1 + sqrt(6) = 2 sqrt(6):

[(2 sqrt(6))/2]^2

(2 sqrt(6))/2 = 2/2×sqrt(6) = [sqrt(6)]^2 - Take its sqrt

= 6

"Annual consumption of bottled water in the United States is given in the table ... etc"

The following should help. Note that I've used a slightly different notation, and I've not shown how I obtained the best-fit values):

(I should add that extrapolating an empirical fit this far from the experimental data is a silly thing to do in general!!)

"Let C be a point not on line AE and D a point on line AE such that CD is perpendicular to AE Meanwhile, B is a point on line CE such that AB is perpendicular to CE.If AB=4, CD=8 and AE=5 then what is the length of CE"

Conservation of momentum:   m₁v₁ + m₂v₂ = 0

Let subscript 1 represent the shot mass, and subscript 2 the gun. m is mass, v is velocity

100*400 + 1000*v₂ = 0  so  v₂ =   -40 m/s      (I assume you meant m/s for velocity, not ms !!)

 Note that the negative sign indicates the gun travels in the opposite direction to the shot.

Assuming constant acceleration (deceleration here) of the gun we can use the constant linear acceleration equation 

v² = u² + 2as  to find the acceleration, a:

0 = (-40)² + 2a*(-0.12)   so a = 1600/0.24  m/s²   (Note: I'm assuming direction is positive in the direction of the shot)

Newton's second law says force = mass*acceleration, so F = 1000*1600/0.24  N

y = 4x - 3    ...(1)

a) If  x  increases by 1 unit, what is the corresponding change in  y? 

Let dy be the change in y. Then   y + dy = 4(x + 1) - 3   or   y + dy = 4x + 1   ...(2)

Subtract equation (1) from equation (2) to find dy.

b) If  x  decreases by 2 units, what is the corresponding change in  y? 

Again let dy be the change in y.  Then  y + dy = 4(x - 2) - 3  or  y + dy = 4x -11   ...(3)

Subtract equation (1) from equation (3) to find dy

.

A ladder 25 feet long is leaning against a wall.
The base of the ladder is sliding away from the wall at 2 feet per second.
Find the rate at which the angle between the ladder and the wall is changing
when the base of the ladder is 7 feet from the wall.

\(\begin{array}{|rcll|} \hline \sin(\phi) &=& \dfrac{x}{25\ \text{feet}} \\\\ \phi &=& \arcsin\left(\dfrac{x}{25\ \text{feet}} \right) \\\\ \dfrac{d\phi}{dx} &=& \dfrac{1} { \sqrt{ 25^2\ \text{feet}^2-x^2} } \quad | \quad \cdot dx \\\\ d\phi &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-x^2} }~dx \quad | \quad :dt \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-x^2} } \dfrac{dx}{dt} \quad | \quad \dfrac{dx}{dt}=2\dfrac{feet}{s},\quad x=7\ \text{feet} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1} { \sqrt{ 625\ \text{feet}^2-7^2\ \text{feet}^2} }\cdot 2\dfrac{feet}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{2}{24\ \text{feet}}\cdot \dfrac{feet}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1}{12}\cdot \dfrac{rad}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{1}{12}\cdot\dfrac{180^{\circ}}{\pi\ \text{rad}} \dfrac{\text{rad}}{s} \\\\ \dfrac{d\phi}{dt} &=& \dfrac{ 15^{\circ}}{\pi s} \\\\ \mathbf{\dfrac{d\phi}{dt}} & \mathbf{=} & \mathbf{ \dfrac{ 4.77464829276^{\circ}}{s} } \\ \hline \end{array} \)

If it is so simple, why are you asking it??

How many 4-digit numbers higher than 4000 can be formed by using the digits 2, 3, 4, 5 and 6 without repetition?

Any that start with 4, 5 or 6 will do.

So that is 3* 4*3*2 = 72 numbers

Thanks Chris,

I get the same answer.  I hope you do not mind Chris, I cannot resist presenting it.

But it is the same as yours :)

\(\frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot \frac{dx}{dt}\\ \frac{d\theta}{dt}=\frac{d\theta}{dx}\cdot 2\\ \)

NOW

\(sin\theta = \frac{x}{25}\\ x=25sin\theta\\ \frac{dx}{d\theta}=25cos\theta\\ When\;\;x=7 \qquad cos\theta = \frac{24}{25}\\ so\\ When\;\;x=7 \qquad \frac{dx}{d\theta}=25\cdot \frac{24}{25}=24\\ so\\ When\;\;x=7 \qquad \frac{d\theta}{dx}=\frac{1}{24}\\ so\\ When\;\;x=7 \qquad \frac{d\theta}{dt}=\frac{1}{24}\cdot 2=\frac{1}{12}\;\;radians/sec\\\)

There are pi radians in 180 degrees

so 1 radian = 180/pi degrees

and   1/12 radians = 180/(12pi) = 4.77 degrees. 

I used this diagram to get the situation when x=7

Write the equation as y = x^-2 + 4x.   

a) Then dy/dx = -2x^-3 + 4  

b) The slope of the tangent line at P is thus  (dy/dx)_tangent = -2(-1)^-3 + 4

The slope of the normal line at P is given by (dy/dx)_normal = -1/(dy/dx)_tangent .  

c)  The tangent line parallel to y = -12x will have slope -12, so will be of the form y = -12x + k, where k is a constant. 

   We must have  -2x^-3 + 4  = -12 for the slopes to match, which gives us the value of x at which this tangent line touches the curve.  Plug this value into y = x^-2 + 4x to find the corresponding value of y.  Plug these values of x and y into  y = -12x + k and rearrange to find k.

I'll leave you to do the actual calculations!

Yes. An informal proof is not difficult to demonstrate.

Numbers that have an odd prime will have at least one sequence of consecutive numbers that returns its sum. Powers of two (2) are the only numbers that do not have odd primes, so there are no consecutive numbers that add to a power of two (2)

Contradiction (proof part 1).

Defining N as the sum of an odd number of consecutive numbers. An odd number of consecutive numbers will have as its average a whole number, that is the middle number. For this, N = a whole number (the average) * odd number (number of consecutive numbers). This means N has an odd number factor.  If \(N = 2^k\) then (N) cannot have an odd number factor. Proof by contradiction that no odd number of consecutive numbers will sum to N, if \(N = 2^k\)

Contradiction (proof part 2)

Defining N as the sum of an even number of consecutive numbers.

An even number of consecutive numbers will have an average that is half the sum of its two middle numbers.

\(N= \frac{(m_1 + m_2)\leftarrow \tiny \text{middle nmbs}}{2} * \text {even # (number of consecutive numbers)}\\ N = (m_1 + m_2) *\frac{1}{2} * \text {even number}\\\)

The sum of any two consecutive numbers is odd. This odd number is a factor of (N).

If \(N=2^k\) then (N) cannot have an odd number factor. Proof by contradiction that no even number of consecutive numbers will sum to N, if \(N = 2^k \)

GA

Here's my take on this:

What is the largest integer less than 2018 that cannot be written as a sum of two or more consecutive integers?

Two or more consecutive integers:

\(\small{ \begin{array}{|r|r|r|l|} \hline \text{consecutive} \\ \text{integers} \\ \hline 2 & n+(n+1) & = 2n + 1 & \text{all odd numbers} \\ \hline 3 & (n-1) + n + (n+1) & = 3n & \text{all multiples of}~3 \\ \hline 4 & (n-1) + n + (n+1)+(n+2) & = 2(2n+1) & \text{all odd numbers} \times 2 \\ \hline 5 & (n-2)+(n-1) + n + (n+1)+(n+2) & = 5n & \text{all multiples of}~5 \\ \hline 6 & & = 3(2n+1) & \text{all odd numbers} \times 3 \\ \hline 7 & & = 7n & \text{all multiples of}~7 \\ \hline 8 & & = 4(2n+1) & \text{all odd numbers} \times 4 \\ \hline 9 & & = 9n & \text{all multiples of}~9 \\ \hline 10 & & = 5(2n+1) & \text{all odd numbers} \times 5 \\ \hline 11 & & = 11n & \text{all multiples of}~11 \\ \hline 12 & & = 6(2n+1) & \text{all odd numbers} \times 6 \\ \hline \ldots & \\ \hline \end{array} }\)

Numbers can be written as a sum of two or more consecutive integers:

\(\begin{array}{|l|} \hline \text{Rule $1$: All odd numbers without $1$ }\\ \text{Rule $2$: All multiples of an odd number without $1$ }\\ \text{Rule $3$: All odd numbers $\times$ an even number without $1$ }\\\\ \text{In general all multiples of an odd number can be written }\\ \text{as sum of two or more consecutive integers (without 1)} \\ \hline \end{array}\)

Example:

\(\begin{array}{|rcll|} \hline 3 & 5 & 7 & 9 & 11 & 13 & 15 & \ldots \\ 6 & 10 & 14 & 18 & 22 & 26 & 30 & \ldots \\ 9 & 15 & 21 & 27 & 33 & 39 & 41 & \ldots \\ 12 & 20 & 28 & 36 & 44 & 52 & 60 & \ldots \\ 15 & 25 & 35 & 45 & 55 & 65 & 75 & \ldots \\ 18 & 30 & 42 & 54 & 66 & 78 & 90 & \ldots \\ 21 & 35 & 49 & 64 & 77 & 91 &105 & \ldots \\ 24 & 40 & 56 & 72 & 99 &104 &120 & \ldots \\ \ldots &\ldots &\ldots &\ldots &\ldots &\ldots &\ldots &\ldots \\ \hline \end{array}\)

So all numbers without an odd prime number in their factorisation cannot be written

as a sum of two or more consecutive integers also the number 1.

Thanks for that answer, PM....

The answers don't have any "space limits" ..........as long as you're giving good answers  [ or.....even attempting a difficult problem..... ] 

However......don't write a book.....the person asking the question will probably not appreciate extremely long answers

Thanks, GA......that's pretty interesting.....

Just a cursory examination of some lower powers of 2  seems to confirm this.....do you know the proof???

Let the side opposite the angle, θ, we are interested in = x

The hypotenuse = 25

So we have

sin θ = x / 25            differentiate both sides with respect to time, t

(d/dt ) sin θ   =     ( d / dt )  ( x /  25)

Remember that  on the left side

d / dt  = d / dθ * dθ / dt        where  d/dθ sinθ = cos θ         

And  on the right side

 d / dt   =  d / dx *  dx / dt          where   d/dx (x/25)  =  1/25

So we have....

 cos θ  * ( dθ / dt)   =  (1/25) * dx / dt            [ note that  dx/dt   = 2ft/s ]

dθ / dt    =     (1 / cos θ)  * (1/25) (2) 

dθ / dt  =   (2 / 25)  sec θ

When the ladder is 7 ft from the wall..... sec θ   =  

25 / √[ 25^2 - 7^2]  =    25 / √[625 - 49 ]  =  25 /  √576    =  25 / 24

So

dθ / dt   =      (2 / 25 )  * ( 25 / 24) rads / sec  =  1 / 12 rads/ sec ≈  4.77° / sec