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An airplane makes a 990 km flight with a tailwind and returns,
flying into the same wind. The total flying time is 3 hours 20 minutes,
and the airplane’s speed in still air is 600 km/h. What is the speed of the wind?

velocity airplane:  \(v_a\)

velocity wind: \(v_w\)

time for the journey there: \(t_1\)

time for way back: \(t_2\)

distance: d = 990 km

total time: \(t = t_1 + t_2\)           

( t = 3 hours 20 minutes )

\(\begin{array}{|rcll|} \hline d &=& (v_a+v_w)\cdot t_1 \quad \text{ for the journey there} \\ d &=& (v_a-v_w)\cdot t_2 \quad \text{ for way back} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline t_1 &=& \dfrac{d}{v_a+v_w} \\ t_2 &=& \dfrac{d}{v_a-v_w} \\\\ t &=& t_1 + t_2 \\\\ t &=& \dfrac{d}{v_a+v_w} + \dfrac{d}{v_a-v_w} \\\\ t &=& d\cdot \left( \dfrac{1}{v_a+v_w} + \dfrac{1}{v_a-v_w} \right) \\\\ \dfrac{t}{d} &=& \dfrac{1}{v_a+v_w} + \dfrac{1}{v_a-v_w} \\\\ \dfrac{t}{d} &=& \dfrac{v_a-v_w+v_a+v_w}{(v_a+v_w)\cdot (v_a-v_w)} \\\\ \dfrac{t}{d} &=& \dfrac{2v_a}{(v_a+v_w)\cdot (v_a-v_w)} \\\\ \dfrac{t}{d} &=& \dfrac{2v_a}{v_a^2-v_w^2} \\\\ \dfrac{d}{t} &=& \dfrac{v_a^2-v_w^2}{2v_a} \quad &| \quad \cdot 2v_a\\ 2v_a\cdot \dfrac{d}{t} &=& v_a^2-v_w^2 \quad &| \quad +v_w^2\\ v_w^2 + 2v_a\cdot \dfrac{d}{t} &=& v_a^2 \quad &| \quad -2v_a\cdot \dfrac{d}{t}\\ v_w^2 &=& v_a^2 -2v_a\cdot \dfrac{d}{t}\\ v_w^2 &=& v_a\cdot \left( v_a - \dfrac{2d}{t} \right) \quad &|d=990 \quad v_a = 600 \quad t=3\frac{1}{3}=\frac{10}{3}\ \text{hours}\\ v_w^2 &=& 600\cdot \left( 600 - \dfrac{2\cdot 990}{\frac{10}{3}} \right) \\ v_w^2 &=& 600\cdot \left( 600 - \dfrac{2\cdot 990\cdot 3}{10} \right) \\ v_w^2 &=& 600\cdot \left( 600 - 6\cdot 99 \right) \\ v_w^2 &=& 600\cdot \left( 600 - 594 \right) \\ v_w^2 &=& 600\cdot 6 \\ v_w^2 &=& 3600 \quad &| \quad \sqrt{} \\ v_w & = & \sqrt{3600} \\ \mathbf{v_w} & \mathbf{=} & \mathbf{60} \\ \hline \end{array}\)

The speed of the wind is \( \mathbf{60 \ \frac{km}{h}}\) .

Hello Guest!

How do i calculate frequency given

inductance and Capacitance?

Thomson oscillation equation

\(F= \frac{1}{2\pi \times \sqrt{LC} } \)

F = resonant frequency in Hertz
L = inductance in Henry
C = capacitance in farads

Greeting asinus :- )  !

10^

how to calculate a=bxc?

a= bxc              [ / bc

\(\frac{a}{bc} = x\)

\(x= \frac{a}{bc} \)

you have to make the denominators the same, in this case they are, so you can just add them together.

-7+3=-4

its the same as 3-7

but you just have to include the denominator so it is -7/4  +  -3/4  =   -4/4  which is -1/1 = -1

hope that helps

Solve the following system:
{3 x-2 y = 7 | (equation 1)
5 x+2 y = 1 | (equation 2)
Swap equation 1 with equation 2:
{5 x+2 y = 1 | (equation 1)
3 x-2 y = 7 | (equation 2)
Subtract 3/5 × (equation 1) from equation 2:
{5 x+2 y = 1 | (equation 1)
0 x-(16 y)/5 = 32/5 | (equation 2)
Multiply equation 2 by 5/16:
{5 x+2 y = 1 | (equation 1)
0 x-y = 2 | (equation 2)
Multiply equation 2 by -1:
{5 x+2 y = 1 | (equation 1)
0 x+y = -2 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{5 x+0 y = 5 | (equation 1)
0 x+y = -2 | (equation 2)
Divide equation 1 by 5:
{x+0 y = 1 | (equation 1)
0 x+y = -2 | (equation 2)
Collect results:
Answer: |x = 1           y = -2

The formula for a convergent infinite geometric series is:  Sum  =  a / (1 - r)

where  a = first term  and  r = common ratio.

This problem has the first term = -3/2   --->   a = -3/2

Since it sums to twice the common ratio,  Sum = 2r.

Putting these together   --->   2r  =  (-3/2) / (1 - r)                        multiply both sides by 1 - r:

                                      --->   2r(1 - r)  =  -3/2                              multiply both sides by 2:

                                      --->   4r(1 - r)  =  -3                                 multiply out the left-hand side:

                                      --->   4r - 4r²  =  -3                                  rearrange the terms:

                                      --->   0  =  4r² - 4r - 3                              factor:

                                      --->   0  =  (2r - 3)(2r + 1)

                                      --->   Either  r = 3/2  or  r = -1/2

Since an infinite geometric series converges only when  -1 < r < r,  the possible answer 3/2 must be rejected.

             --->   answer:  r = -1/2 

Ok but what I need is to make an equation that has only x as a variable that will =

log (  (1000 (square root of (A^3) ) ) / A^-1 )

Cause right now if we solve in terms of "a", like you said, wouldn't it leave me with a new undefined variable?

like that "a" would be = the equation with only "x" as variable that is needed.

Thanks a lot! :)

Hehe well by hand x= -2 gives 0 .

Seems the calculator fails with the parenthesis or something. My bad...

Thanks a lot guys.

I'm trying to work this out with gauss and it doesn't add up.

2x^2 + 3x - 2 's rationnal roots can be - or + 1,  - or + 2. 

I try replacing x with all four options but nothing =0. Is this normal?

You can graph here, but it's clunky......I prefer Desmos......here's the website.......

https://www.desmos.com/calculator

(2x^2 + 3x - 2 )   / ( 6x+12)

[(2x - 1) (x +2)] /  [ 6 (x + 2) ]         the (x + 2)  factors "cancel"

(2x - 1) / 6

Simplify the following:
(2 x^2+3 x-2)/(6 x+12)

Factor 6 out of 6 x+12:
(2 x^2+3 x-2)/6 (x+2)

Factor the quadratic 2 x^2+3 x-2. The coefficient of x^2 is 2 and the constant term is -2. The product of 2 and -2 is -4. The factors of -4 which sum to 3 are -1 and 4. So 2 x^2+3 x-2 = 2 x^2+4 x-x-2 = 2 (2 x-1)+x (2 x-1):
2 (2 x-1)+x (2 x-1)/(6 (x+2))

Factor 2 x-1 from 2 (2 x-1)+x (2 x-1):
(2 x-1) (x+2)/(6 (x+2))

((2 x-1) (x+2))/(6 (x+2)) = (x+2)/(x+2)×(2 x-1)/6 = (2 x-1)/6:
Answer: |(2x - 1)/6

(27*-158)*6+7*0.9^2+(48-0.5)-58/0+1457*(7-7)  =

(27*-158)*6+ (7*0.9^2) +(48-0.5)-58/0+1457*(0) =

(27*-158)*6+7*0.9^2+(48-0.5)-58/0  =

(-4266)*6  + 5.67 + (47.5) - (58/0)  =

-25596 + 5.67 + 47.5   - (58/0)   = 

-25542.83  -  (58/0)

Notice that, at this point, we have a problem with the last term......since division by 0 is undefined, then the final result is also undefined........

Hello Guest!

Find all numbers for which the

rational expression is not defined.

4x / (x² - 2x - 15)

The rational expression is undefined

when the denominator is zero.

x² - 2x - 15 = 0

\(x = {2 \pm \sqrt{4+60} \over 2}\)

\(x= \frac{2\pm 8}{2} \)

\(x_{1}= 5\ \ x_{2}= - 3 \)

The rational expression is undefined when

\(x\in \left\{5, -3 \right\} \)

Greeting asinus :- )  !

4x /(x^2-2x-15)        factor the denominator as  (x - 5)(x +3)

And since anything that makes the denominator 0, is undefined......then this rational function will be undefined when x = 5 or x = -3

See the graph, here: https://www.desmos.com/calculator/81t61hpgz7

If you set it up to =a, for example, then you will have this solution:

log (  (1000(square root of ( (10^x)^3) ) ) /(10^x)^-1)=a, solve for x

x = (2 (a - 3))/5

(sqrt of 125)^3x = 625^(2x-1)

Tony, we can solve this one by equatiing bases.......notice that.....625 = 5^4

And sqrt of 125   can be wriiten as  [125]^(1/2) =  [ (5)^3]^(1/2)    =  5^(3/2)

So we have

[ 5^(3/2) ] ^(3x)   =  [5^4 ] ^ (2x - 1)     simplify

5^ [ ( 9/2 ) x] =   5^ [ 8x - 4]

Since we have the bases the same, we can just solve the exponential equations....so we have

(9/2)x = 8x - 4        multiply through by 2

9x = 16x - 8        add 8 to both sides and subtract 9x from both sides

8 = 7x               divide both sides by 7

8 / 7  = x

Well, you can specify what x equals beforehand and you will get an answer. You can also have the expression equal to another variable  such as "a". Then you can solve for x in terms of "a".

Take a good look at the line before that. What is it you don't understand?? If you have:

a^b=c^d, then take the log of both sides and you have:

b*log(a)=d*log(c). Do you get it?