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Force = Mass * Acceleration

This equation is missing the time taken. With the time taken s. We can work out the acceleration. Since mass is equal to 1. Force = Acceleration (In this case only)

Acceleration: Speed / time which translate to 1 / s^2

I think :P

"what is real and imaginary part in sin(x+iy)? and its solution?"

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What's the mean?  What's the standard deviation?  What's the distribution?!!

We need more information than you have provided.

In a shop, the cost of 4 shirts, 4 pairs of trousers and 2 hats is $560. The cost of 9 shirts, 9 pairs of trousers and 6 hats is $1,290. What is the total cost of 1 shirt, 1 pair of trousers and 1 hat? 

It might be difficult to solve this outright with a system of equations because we might have more variables than equations.....let's see if we might arrive at some logical conclusion...

Note that  if I divided the second thing by 3, we would have that

3 shirts, 3 pairs of trousers and 2 hats is $430

And taking the first we have that 4 shirts, 4 pairs of trousers and 2 hats is $560

Then. since the cost of the hats is the same, it must be that   1 shirt and 1 pair of trousers  cost

 $560 - $430  = $130

So....9 sets of  each of them must cost  9 * $130 = $1170

So....the cost of a hat =  [ 1290 - 1170] / 6  = 120/6  = $20

So....the total cost of 1 shirt, 1 pair of trousers and 1 hat =  $130 + $20 = $150

Check  

4(130) + 2(20)  = $560

9(130) + 6(20) = $1290

Four children have small toys. The first child has 1/10 of the toys, the second child has 12 more toys than the first, the third child has one more toy of what the first child has and the fourth child has double the third child. How many toys are there? 

Let's call the total number of toys  " t " .

first child's toys  =  t/10

second child's toys  =  12 + first child's toys  =  12 + t/10

third child's toys  =  1 + first child's toys  =  1 + t/10

fourth child's toys  =  2 * third child's toys  =  2 * (1 + t/10)  =  2 + 2t/10

So.....

t  =  first child's toys + second child's toys + third child's toys + fourth child's toys

t  =  ( t/10 ) + ( 12 + t/10 ) + ( 1 + t/10 ) + ( 2 + 2t/10 )

t  =  t/10 + 12 + t/10 + 1 + t/10 + 2 + 2t/10                     Combine like terms

t  =  5t/10 + 15

t  =  t/2 + 15                Subtract  t/2  from both sides.

t - t/2  =  15

t/2  =  15                     Multiply both sides by  2  .

t =  30

 (81^-.25)^3

Note that this is really the same as

[ 81 ^ (-1/4)] ^3  =

[ 1/ 81^(1/4) ] ^3  =

[1 / ⁴√81 ]^3  = 

[ 1 / 3 ]^3  =  

1 / 27       .....just as PsychoBaker found....!!!

For what value of x will the function f(x) = -3(x - 10)(x - 4) have a maximum value? Find the maximum value.

We could do some fancy calculations here, but note that this is an "upside-down" parabola that has zeroes at   x = 4  and at x = 10

And, because of symmetry, the max value will occur halfway between these, i.e., at x = 7

And the max value will be  -3(x - 10)(x - 4) =  -3(7 - 10) (7 - 4)  =  -3 * -3 * 3  =  27

A class average mark in an exam is 70. The average of students who scored below 60 is 50. The average of students who scored 60 or more is 75. If the total number of students in this class is 20, how many students scored below 

Note that if the class average is 70, then the total number of points scored is  70 * 20  = 1400

Call the number of students who scored below 60, x .....So....their total number of points they scored is just their average times their number =  50x

Call the number of students who scored above 60, 20-x .....So....their total number of points they scored is just their average times their number =  75 (20 - x)

So.....we have this equation :

50x + 75 (20 - x)  = 1400       simplify

50x + 1500 - 75x  = 1400      subtract 1500 from both sides

-25x = - 100      divide both sides by -25

x = 4  ..so.....4 scored below 60

how did you get 1/3? :)

1.

The question is...

$8,880  is  6%  of what number?               Let's call the number  " p " for selling price.

$8,880  =  0.06 * p                                      Divide both sides by 0.06

$8,880 / 0.06  =  p

$148,000  =  p

(81^-.25) = 1/3. 1/3^3 = 1/3 * 1/3 * 1/3 = 1/27. 

The question is...

1300  out of  3000  is  what  out of  100

\(\frac{1300}{3000}\,=\,\frac{\text{what}}{100} \\~\\ 0.4\overline3\,=\,\frac{\text{what}}{100} \\~\\ 0.4\overline3\,*100=\,\text{what} \\~\\ 43.\overline3\,=\,\text{what}\)

So...

1300  out of  3000  is  43\(\frac13\)  out of  100  .

1300  out of  3000  is  43\(\frac13\)  %

Here's a direct way using the change-of-base method

log_b (x) = (3/4) log₁₀ (x)

log (x) / log (b)  = (3/4) log (x) / log 10        divide both sides by log (x)  and note that log10  = 1

1/ log (b)  = 3/4       which implies that

log (b)  = 4/3    which implies that

10 ^ (4/3)  = b

[10^4] ^(1/3) = b

10000^(1/3) = b = ∛10000

9,250,000,000,000,000,000,000

BTW....it is easy to prove that, at a minimum, at least one point "P" exists

Let the y coordinate of our  P  = 6

And using the equation for AB, y = √3x,   and the formula for the distance from a point to a line....let the distance from P to AB = 5.....so we have 

y = √3x   →  √3x - y  =   0

abs ( Ax + By + C) / √ [ A^2 + B^2 ]       where A = √3, B = -1   and  C = 0

l √3x - 1(6) l /  2  = 5 → [√3x - 1(6)] = 10 → √3x = 16 →  x = 16 / √3

So...given the specified altitudes to the respective sides,  the value for P  =

 ( 16 / √3,  6 )

Here's the approximate pic with the three triangles APB, BPC and APC....and  EP, FP and GP are the required altitudes to each side 

If I understand the problem correctly, there will be three triangles formed, each with an equal base  which will be  a side length of ABC and respective altitudes of 5,6  and 7

So.....the total  area of these three triangles can be represented as :

(1/2) s ( 5 + 6 + 7 )  = (1/2) s ( `18)  =  9s    (1) 

Where s is the side length of the equilateral triangle ABC

And the area of an equilateral triangle can be represented as

(1/2) s^2 ( √3 / 2 )  =  (√3 / 4) s^2        (2)

Set (1)  = (2)  and solve for s

9s =  (√3 / 4) s^2      divide by s

9 = (√3 / 4) s     multiply both sides by  4/ √3

9 *  4 / √3  =  s

36 / √3  =  s  =   12√3 

So......using (1)...... the area of ABC  is   9 (12√3)  =  108√3  sq units

The guest above me gave you the method, but you mentioned the least common denominator of 3 or more fractions.

i think i can assume you know how to find the least common denominator for 2 numbers. So you could also do this:

Suppose you need to find the least common denominator for {a₁, a₂, ...., a_n}.

First step: find the least common denominator for a₁ and a₂.

Replace a₁ and a₂ with x₁, their least common denominator.

Second step: find the least common denominator for x₁ and a₃. 

Replace x₁ and a₃ with x₂, their least common denominator.

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Continue this until you get to x_n-1.

I know the guest above me already answered the question, i just wanted to show you you can use what you already know to solve this question ;).

I'm assuming that the last one is :

 2 / (x+2)   ≤   x / (x+1)

Setting each factor to 0 and solving for x, the critical values are  -2  and - 1

Note that the inequality is undefined at these two x values

Turn this into an equality

Cross-multiplying, we have

2 (x + 1)  =  x ( x + 2)     simplify

2x + 2  =  x^2 + 2x     subtract 2x from both sides

2  = x^2     rearranging, we have

x^2 - 2 = 0

And this will be true when   x = - √2   or   x = √2

And these are the other two critical values that we need

Note that the inequality will  be true when

x < -2 ,     - √2   ≤ x < -1   and  x ≥ √2

Here's a graph :

https://www.desmos.com/calculator/zu0vmqfgux

Mathematica 11 gives the following as "Multiple-argument Formulas", but no step by step breakdown:

Cos[Pi/7] + Cos[Pi/7]^2 - 2 Cos[Pi/7]^3 == -1 + 2 Cos[Pi/14]^2 + (-1 + 2 Cos[Pi/14]^2)^2 - 2 (-1 + 2 Cos[Pi/14]^2)^3

Cos[Pi/7] + Cos[Pi/7]^2 - 2 Cos[Pi/7]^3 == 1 - 2 Sin[Pi/14]^2 + (1 - 2 Sin[Pi/14]^2)^2 - 2 (1 - 2 Sin[Pi/14]^2)^3

Cos[Pi/7] + Cos[Pi/7]^2 - 2 Cos[Pi/7]^3 == -3 Cos[Pi/21] + 4 Cos[Pi/21]^3 + (-3 Cos[Pi/21] + 4 Cos[Pi/21]^3)^2 - 2 (-3 Cos[Pi/21] + 4 Cos[Pi/21]^3)^3

I'm assuming that we have :

2.)   (3x-1) (2-5x) / (2x+1) > 0

Setting each factor to 0 and solving for x, the critical values are :

x = 1/3 , x = 2/5  and x = -1/2

The function will be undefined when x = - 1 /2

The numerator is positive when    1/3 < x < 2/5

And the denominator is positive when x > -1/2  

And the function is > 0 whenever the numerator and denominator have the same sign

So...the function is > 0  when

1/3 < x < 2/5       or when     x < -1/2

Here's the graph :

https://www.desmos.com/calculator/utup8jcqp2

I'm assuming this is

1.)   (x^3+27) / (x^2-3x-10) < 0     factoring the numerator and denominator, we have

[(x + 3) (x^2  - 3x + 9)] / [ ( x - 5) (x + 2) ] < 0

The second factor in the numerator is always positive so the numerator is < 0 whenever x < -3

The denominator is undefined whenever x = 5 or x = -2

And the denominator will < 0  whenever    -2 < x < 5

So......the function will be < 0 whenever the numerator is positive and the denominator is negative  or when  the numerator is negative and the denominator is positive

The first situation will occur when    -2 < x < 5

And the second wil occur when    x < - 3

So the solution set is  [  x < - 3 ]    U   [    -2 < x < 5  ]

Here's the graph :

https://www.desmos.com/calculator/xgh0b1qmvm

12 x 1/3 = 4

4 + 6 = 10 Cucumber plants that Carlos planted this year.

WolframAlpha tells me the answer is 1/4 but I'd really like someone to tell me why ://

I liked our guests logic so I decided to recount to work out where I went wrong.

This is what I found :)

I double counted for 2 in one and 2 in another, which meant my answer was 18 too big.

Double counting is a common mistake with these typers of questions which is why it is better to use logic like our guests if you can..

Thanks Guest - I am really pleased that you showed me your straight forward way to do it :)