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Possibly find three fractions with a numerator of one that will add up to 25/27 (problem one) and 15/16 (problem two). I tried 15/16 quickly as it seems this might be a nice challenge unless I'm over thinking it. (1/2)+(1/4)+(1/8)= 14/16 which is off from 15/16

VIETNAMESE??

f'(x)= 1/3x3+ 4x 

f'(x)= (1/3)x^3+ 4x 

Is this what you mean?   The symbol for 'to the power of' is ^

\(f'(x)= \frac{1}{3}x^3+ 4x \\ f(x)=\int\;\frac{1}{3}x^3+ 4x\;dx \\ f(x)=\frac{1}{3*4}x^4+ \frac{4x^2}{2}+c\\ f(x)=\frac{x^4}{12}+ 2x^2+c\\\)

Two fractions, two questions ?

(Can't do either).

BTW, what's the language ?

This is the Google translation:

Write 25/27, 15/16 as the sum of three fractions with different denominators, and have a numerator of 1.

But I still do not understand        Why are there 2 fractions ?  Anyone got any thoughts?

please help!!!

What do you mean by  " 1 = 97 "  ?

If  x = 3  and  y = 97  , then to find

    5x - 6y          , replace  " x "  with  3  and replace  " y "  with  97  .

=  5(3) - 6(97)

=  15  -  582

=  -567

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How far away would light have to be from a black hole with the mass of the Earth (6 x 10²⁴ kg)

in order to be able to escape from its gravitational pull?

(Note: use the formula for escape speed again: vesc = 1.15 × 10-5 MR−−√MR )

Let speed of light \(c = 3\cdot 10^8\ \frac{m}{s}\)

Let gravitational constant \( = G \) \( \)

Let Radius = R

\(\begin{array}{|rcll|} \hline v_{esc} &=& \sqrt{\frac{2 G M_{\text{Earth}} }{R} } \\ v_{esc} &=& \sqrt{2 G} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \quad & | \quad \sqrt{2 G} \approx 1.15 \cdot 10^{-5} \\ v_{esc} &=& 1.15 \cdot 10^{-5} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \quad & | \quad v_{esc} = c \\ c &=& 1.15 \cdot 10^{-5} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \\ c^2 &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ M_{\text{Earth}} }{R} \\ R &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ M_{\text{Earth}} }{c^2} \quad & | \quad M_{\text{Earth}} = 6 \cdot 10^{24}\ kg \qquad c = 3\cdot 10^8\ \frac{m}{s} \\ &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ 6 \cdot 10^{24} }{(3\cdot 10^8)^2} \\ &=& 1.15^2 \cdot 10^{-10} \cdot \frac{ 6 \cdot 10^{24} }{ 9\cdot 10^{16} } \\ &=& 1.15^2 \cdot 10^{-10} \cdot \frac{2}{3} \cdot 10^{(24-16)} \\ &=& 1.15^2 \cdot \frac{2}{3} \cdot 10^{(24-16-10)} \\ &=& 0.88167 \cdot 10^{-2} \ m \\ &=& 8.8167 \cdot 10^{-3} \ m \\ &=& 8.817 \ mm \\\\ \mathbf{R} & \mathbf{ \approx }& \mathbf{ 8.8 \ mm } \\ \hline \end{array} \)

The light would have to be more than \(\mathbf{8.8 \ mm}\) from the black hole with the mass of the Earth

in order to be able to escape from its gravitational pull

Find the ratio of the area of triangle BCX to the area of triangle ACX in the diagram if CX bisects the angle ACB. 

Express your answer as a common fraction.

Let \(A_1=\) area of BCX

Let \(A_2=\) area of ACX

Let \(\varphi = \angle ACB\)

Let ratio = \(\frac{A_1}{A_2}\)

\(\begin{array}{|lrcll|} \hline (1) & 2\cdot A_1 &=& \overline{AX}\cdot 27 \cdot \sin(\frac{\varphi}{2}) \\ (2) & 2\cdot A_2 &=& \overline{AX}\cdot 30 \cdot \sin(\frac{\varphi}{2}) \\ \hline \frac{(1)}{(2)}: & \dfrac{2\cdot A_1}{2\cdot A_2} &=& \dfrac{\overline{AX}\cdot 27 \cdot \sin(\frac{\varphi}{2}) }{\overline{AX}\cdot 30 \cdot \sin(\frac{\varphi}{2}) } \\ & \dfrac{ A_1}{ A_2} &=& \dfrac{ 27 }{ 30 } \\ & &=& \dfrac{ 3\cdot 9 }{ 3\cdot 10 } \\ & &=& \dfrac{ 9 }{ 10 } \\\\ &\mathbf{ ratio } & \mathbf{=} & \mathbf{ \dfrac{9}{10} } \\ \hline \end{array}\)

10 + y/3  = 6      subtract 10 from both sides

y/3  =  -4           multiply both sides by 3

y  =  -12

What are you asking here?

By Euclid, whenever an apex angle is bisected, we have the following relationship :

AX/AC  = BX/BC

AX/30  = BX/27   which implies that

27/30  = BX/AX 

9 /10  = BX/AX     →  BX = (9/10) AX

And triangle ACX  has the same height  as triangle BCX 

And again, by Euclid, triangles under the same height are to each other as their bases

So....the area of triangle BCX = (9/10)ACX

And the ratio of their areas is  :   

BCX :  ACX  =   9 : 10

I don't know what your formula is supposed to calculate! I know it is about calculating the "escape velocity", but I don't need it here! You should know that if the Earth was squeezed into a black hole, the black hole would have a radius of 9 millimeters! The "photon sphere", or the orbit of light would be: 1.5 x 9 =13.5 mm from the center of the black hole. Or, 13.5 - 9 =4.5mm from the "event horizon". That is the best I can do!.

Actually, another method involves knowing relationships of right triangles. This particular triangle is an isosceles right triangle. It's an isosceles triangle because of the tick marks in the diagram. By the definition of the isosceles triangle theorem, the angles opposite the congruent sides are congruent.

However, we already know one of the sides is a right angle, or 90 degrees. Let's solve for the missing angles:

Let alpha = the measure of the unknown angle

Look at that! The missing angles are both 45 degrees? Is that useful? Yes, it is, actually. Because we have proven that the other angles are 45 degrees, we can utilize the 45-45-90 triangle relationships!

Remember that in a 45-45-90 triangle, the ratio of the legs to the legs to the hypotenuse is \(1:1:\sqrt{2}\). 

Now, knowing this, let's figure out x.

I seem to like this method better because it does not require nearl as much work, but both are valid methods of solving. However, I would agree that it is easy to default to Pythagorean Theorem in right triangles. 

200*.05*8/12=

\( \color{red}\frac{200\cdot 0.05\cdot 8}{12}\\ =\frac{2^4\cdot 5}{2^2\cdot3}\\ =\frac{2^2\cdot 5}{3}\)

\(\frac{200\cdot 0.05\cdot 8}{12}=6\frac{2}{3}\)

  !

Thanks, heureka, for deriving that one, step-by-step.....

It's certainly not intuitive that the final result will be "1".....!!!!!

Going from Max's answer, we have

sin 2θ / [2 (1 + sinθ)] 

2sinθcosθ / [2 ( 1 + sin θ) ]

sinθcosθ / (1 + sinθ)

sinθcosθ ( 1 - sinθ) / [1 - sin^2θ] 

sinθcosθ [ 1 - sinθ] / cos^2θ

sinθ[1 - sinθ]/cosθ 

sinθ/cosθ  - sin^2θ/cosθ

tanθ - tanθsinθ

The moral here is that with these trig expressions, there may not be just one "correct" answer !!!!

sin( 2pi )  =  0

thanks!

Since the unlabeled side is the same length as  x , we can also label it  " x " .

Because this is a right triangle, we can use the Pythagorean theorem to find the side lengths...which says...

(leg)² + (other leg)²  =  (hypotenuse)²       Plug in the information from our problem.

x²  +  x²   =   a²                                         Combine like terms.

2x² =  a²                                                    Divide both sides of this equation by  2  .

x²  =  a² / 2                                                Take the positive square root of both sides.

x   =   √[ a² / 2 ]

x   =   a /√2                                                 Replace  " a "  with  " 8.1 " .

x   =   8.1 / √2

x   ≈   5.73   cm

Notice that the term in which the denominator ends in 2013 [1/1+2+3....+2013] is actually the 2012th term. Similarly, the term in which the denominator ends in 2014 [1/1+2+3.....+2014] is actually the 2013th term. Therefore, it is only necessary to sum up the terms 1 to 2013: Sum{2/(n^2 + 3n +2)}, from n=1 to 2013=2013/2015 + 2/2015 = 1.

\(\frac{\frac{2}{5}}{\frac{4}{7}}=\frac{2*7}{4*5}=\frac{14}{20}=\frac{7}{10}=0.7\)

Go online and use this interpolation calculator and try various scenarios until you get your rate:

http://www.ajdesigner.com/phpinterpolation/linear_interpolation_equation.php#ajscroll

Thanks :)

This site does look interesting :)