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sqrt (160)  =   sqrt (16 *10)   =  4sqrt 10

sqrt (252)  =  sqrt (36 * 7)  =   6sqrt 7

sqrt (245)   =   sqrt ( 49 * 5)  =  7sqrt 5

sqrt (180)  = sqrt (36 * 5)  = 6 sqrt (5)

So  we  have

4sqrt (10)            7 sqrt (5)                 28sqrt (10)             

_________  *   __________    =     ____________   = 

6  sqrt (7)            6 sqrt (5)                36 sqrt (7)

(7/9) sqrt ( 10) / sqrt (7)  =

(7/9) sqrt (70)  / 7  =

sqrt ( 70)  *  7/63   =

sqrt (70)  /  9

1/8  + 1/b  = 1/2

1/b = 3/8

b = 8/3 hr

Let Benjamin's  rate per hour   = 1/x

(1/8  + 1/x)   =     ( x + 8)  /  (8x)

Invert  the  right side  and set it  =   2

So

8x / ( 8 + x)  =   2

8x  = 2 ( 8 + x)

8x  = 16 + 2x

8x  - 2x  = 16

6x  = 16

x   =  16/6   =  8/3 hrs.   ≈  2.7  hours  working on his own

(1  + 5)x^3  =   6x^3

Assume   that  the points  are  in  this order   XZY

XZ  =   ZY

So

X =    [  -1 -  ( 1 - - 1)   ,  -7 - ( -7 - - 7) ]   =  [ -1-2 , -7 - 0 ]  =  ( -3, -7)

Sum of coordinates =   -10

Radius is 12.

Assuming a monic quadratic polynomial ....we  must  have  the point  (5, k)  as  the  vertex

Since  (-8, -7)  is on the  graph, we  have  that

-7  = ( -8 - 5)^2 +  k

-7  = (-13)^2  + k

-7  = 169 + k

k =  -7 -169  =  -176

So   the  function is

y=  (x - 5)^2  - 176

So......the point (5, -176)  also must be on the graph

x and x +2   are NOT exactly consecutive integers......they are either consecutive ODD or EVEN integers

x and x+1 are consecutive integers     so I am not sure what your question means.....but if it is supposed to be x and x+1 then: 

1/x   +  1/(x+1) = 29/420   multiply both sides by x(x+1)

x+1  + x  = 29/240 ( x^2 + x)       now solve for 'x'                                then the other number is x+1 

If your Q IS supposed to be x and x+2 then

1/x + 1/(x+2) = 29/240     multiply both sides by x(x+2)

x+2    + x    = 29 / 240 ( x^2 + 2x)       solve for x                               then the other number is  x+2

Can you finish?

I used a 3-4-5 right-angled triangle.  

(Btw, the above answer's mine.) 

Though a question..:
you referred to "v" as velocity, shouldn't it be speed?

(Because K.E. is scalar not vector, and scalar quantities must be of either: Vector x Vector or scalar x scalar). 

Oh, that must be the correct answer! Thanks for explaining this idea about mass doubling due to time being considered! (Never thought of it before)!
Sorry guys for my completely wrong answer(guest) above. If u can delete it somehow go ahead as I can't.

As shown in the diagram,\(~BD/DC=2,~CE/EA=3\),  and \(AF/FB=4\).
Find \([DEF]/[ABC]\).

\(\begin{array}{rcll} \text{ Let } BC&=&a\\ \text{ Let } BD&=&\frac{2}{3}a \\ \text{ Let } DC&=&\frac{1}{3}a \\\\ \text{ Let } CA&=&b \\ \text{ Let } CE&=&\frac{3}{4}b \\ \text{ Let } EA&=&\frac{1}{4}b \\\\ \text{ Let } AB&=&c \\ \text{ Let } AF&=&\frac{4}{5}c \\ \text{ Let } FB&=&\frac{1}{5}c \\ \end{array}\)

\(\begin{array}{rcll} \text{ Let } \angle CAB&=& A\\ \text{ Let } \angle ABC&=& B\\ \text{ Let } \angle BCA&=& C\\ \end{array}\)

\(\begin{array}{|rcll|} \hline 2*[ABC] &=& bc\sin(A) \\\\ 2*[AFE] &=& \frac{1}{4}b*\frac{4}{5}c \sin(A) \\ 2*[AFE] &=& \frac{1}{5}bc \sin(A) \\ 2*5*[AFE] &=& bc \sin(A) \\ \hline bc\sin(A) = 2*[ABC] &=& 2*5*[AFE] \\ 2*[ABC] &=& 2*5*[AFE] \\ [ABC] &=& 5*[AFE] \\ \mathbf{[AFE]} &=& \mathbf{\frac{1}{5}[ABC]} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline 2*[ABC] &=& ca\sin(B) \\\\ 2*[FBD] &=& \frac{1}{5}c*\frac{2}{3}a \sin(B) \\ 2*[FBD] &=& \frac{2}{15}ca \sin(B) \\ 2*\frac{15}{2}*[FBD] &=& ca \sin(B) \\ \hline ca \sin(B) = 2*[ABC]&=&2*\frac{15}{2}[FBD] \\ 2*[ABC]&=&2*\frac{15}{2}[FBD] \\ [ABC]&=&\frac{15}{2}[FBD] \\ \mathbf{[FBD]} &=& \mathbf{\frac{2}{15}[ABC]} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline 2*[ABC] &=& ab\sin(C) \\\\ 2*[EDC] &=& \frac{1}{3}a*\frac{3}{4}b \sin(C) \\ 2*[EDC] &=& \frac{1}{4}ab \sin(C) \\ 2*4*[EDC] &=& ab \sin(C) \\ \hline ab \sin(C) = 2*[ABC] &=& 2*4*[EDC] \\ 2*[ABC] &=& 2*4*[EDC] \\ [ABC] &=& 4*[EDC] \\ \mathbf{[EDC]} &=& \mathbf{\frac{1}{4}[ABC]} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline [DEF] + [AFE]+[FBD]+[EDC] &=& [ABC] \\ [DEF] + \frac{1}{5}[ABC]+\frac{2}{15}[ABC]+\frac{1}{4}[ABC] &=& [ABC] \\ [DEF] + [ABC] \left(\frac{1}{5}+\frac{2}{15}+\frac{1}{4} \right) &=& [ABC] \\ [DEF] + \frac{7}{12}[ABC] &=& [ABC] \\ [DEF] &=& [ABC]- \frac{7}{12}[ABC] \\ [DEF] &=& [ABC] \left(1- \frac{7}{12} \right) \\ [DEF] &=& \frac{5}{12}[ABC] \\ \mathbf{\frac{[DEF]}{[ABC]}} &=& \mathbf{\frac{5}{12}} \\ \hline \end{array}\)

Let X, Y, and Z be points such that If XZ/XY = ZY/XY = 1/2.  If Y = (1,-7) and Z = (-1,-7), then what is the sum of the coordinates of X?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Y=(1, -7)       Z=(-1, -7)

X = (-3, -7)

Sum of coordinates of X = -10

re-write the second equation to

3a + 4b = k           now, if the two equations are equal, there are infinite solutions....   so   k = 7

x+y = 12     or     y = 12-x

1/x - 1/y = 2/9       sub in the red equation

1/x  - 1/ (12-x) = 2/9     now solve for 'x'       then y = 12-x

https://web2.0calc.com/questions/june-cleaver-needs-2-000-in-7-years-time-so-she-can#r1

You already posted this....... check the earlier post....don't be lazy!

Area of a rectangle = length * width

a) width = area / length =   (2x^2-x-1) / (2x+1)

b) non-permissable values would be values that result in a denominator of 0      2x+1 = 0     x  cannot equal -1/2

c) Length cannot be LESS than zero either,   soooo    2x+1 cannot be less than 0

          2x+1 < 0

            x < -1/2     x cannot be less than -1/2

Here you go guest:

if m² + 1 = 3m

we can simplify like this:

(m⁵+2m³-9m+27)/3m

= (m³(m²+2)-9m+27)/3m

= (m³(3m+1)-9m+27)/3m

= (3m⁴+(m³-9m)+27)/3m

= (3m⁴+m(m²-9)+27)/3m

= (3m⁴+m(3m+10)+27)/3m

= (3m⁴+3m²+10m+27)/3m

= (3m²(m²+1)+10m+27)/3m

= (3m²(3m)+10m+27)/3m

= (9m³+10m+27)/3m

= (m(9m²+10)+27)/3m

= (m(27m+1)+27)/3m

= (27m²+m+27)/3m

= (27(m²+1)+m)/3m

= (27(3m)+m)/3m

= (81m+m)/3m

=(82m)/(3m)

= 82/3

= 27 1/3

P.S. I probably made a mistake somewhere, but this is the general idea of how to do it!

Here you go guest.

If this is the graph of f(x)

f(-2) would mean the y-coordinate when x = 2

SO the answer is (2,-2) for the coordinate

SO the answer is 

-2

Looka at the graph.....look at the line x = 2     find where it intercepts the graph at y = -2     this is f(2)

The vertex of the parabola will be shifted 3 units to the LEFT  to   -1,-1

(I'm assuming that the 2 k's are different; for simplicity purposes, I'm gonna call them 3a+1 and 6b+1)

We can prove this statement by contradiction. 

Notice that if a is even, we can represent it as 2(3b)+1=6b+1 for some integer b. Let's say that a is not even.

If a is odd, 3a will also be odd, but if you add one to an odd number, it automatically becomes even, meaning that 3a+1 will always be even for odd values of a. If a number is even, it will always be composite because it has a factor of 2 (except if it is 2 itself, but it's irrelevant since it's unobtainable). Therefore, a must be even, meaning that it must be expressible as 6b+1 for some integer b.

9 % annual interest  compounded monthly is   .09/12 = .0075 %   periodic interest

7 years is  7 * 12 = 84 periods

2000 is future value

2000 = d * ( 1+.0075)⁸⁴        where d = initial deposit amount

d = 1067.69 dollars initial deposit

Neither of you two dumb-dumbs have the correct answer.

The change in power is proportional to the ratio of the cubed velocities.

$\Delta P = \dfrac {V_2^3}{V_1^3}$

$\Delta P = \dfrac {10^3}{5^3} = 8$

Then

$0.5MW *8 = 4 MW$

This is solvable using kinetic energy equations. But you have two errors in your second equation.

Your second equation has an erroneous value for the mass. Also, you multiplied by the coefficient of efficiency instead of dividing. In this case, you should have done neither.

Dividing gives the actual kinetic energy of the air that’s required to produce this power output.  However, if you do this on the second equation you have to do it on the first equation. The results will be the same if the proportions are the same. Despite its mention in the question, the coefficient is not needed to solve this question. In fact, most of the BS written in the context isn’t needed for the question.      

$E_k=\dfrac{1}{2}(40000)(10^2) = 2.0MJ$ – Your result without multiplying by the coefficient of efficiency.

Consider carefully: if you have twice the velocity there will also be twice the mass in the same unit of time.

So...

$E_k=\dfrac{1}{2}(80000)(10^2) = 4.0MJ$

------------

Defining how kenetic energy relates to power:

$KE = \dfrac{1}{2} M * v^2$ the velocity may be expressed as $\left(\dfrac{\dfrac{m}{s}}{s}\right)$

And power as:

$P = \dfrac{KE}{\Delta t} | \;\; t \;\; \text { is in seconds.}$

Then

$P = \dfrac{1}{2}*\left(\dfrac{{\dfrac{\dfrac{m}{s}}{s}}}{s}\right)$

Then

$P = 0.5 M * v^3$

------------

A point about your hint:

Hint: (Although the text is talking about power and energy, they are the same here due to "1 second" (Since P=E/t -->P=E if t=1))

Power and energy are related and one can be derived from the other, but they are never the same. Just as kilometers per hour is never the same as kilometers. It does not matter that the time unit is one (1); they have different units: for power it’s watts and Joules per second. For energy it’s Joules and watt-seconds.

------------

The general equation for power produced by a wind turbine is

$P = 1/2 \pi * r^2 * v^3 * \rho * \eta  \;\;| \;\; \tiny \text {where r is the radius of the turbine blade, v is velocity of the wind, $\rho$ is the air density, and $\eta$ is the efficiency factor.}$ 

With this equation you can reconstruct the basic parameters of the wind turbine. 

This equation was not included, but a lot of BS filler was.

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Both the question and the guest answer look like they could have been written with a quill...

That is, written in a narrative style consistent with the lecture circuit of the nineteenth century.    

The great masters of physics and science, such as Michael Faraday and Joseph Henry gave public lectures on various subjects. But for every true master there were dozens of morons, idiots and charlatans, who either did know the subject or couldn’t teach it with any kind of coherent logic.   

Here’s an example in written form.