Answer:
Step-by-step explanation:
i got you luv
whats up?
Answer:
x = 13
Step-by-step explanation:
No se exactamente como decir esto porque yo ayudo la majoria de los usuario en ingles pero hay que usar un método para describir a x
Un método que se llama Pythagorean theorem
Pythagorean theorem: a^2 + b^2 = c^2 donde a^2 = 5 b^2 = 12 y c^2 = x
Entonces…
5^2 + 12^2 = c^2
Nota: podemos ver que esto esta bien porque el sum de los dos lados suma el mas largo.
Ahora
25 + 144 = c^2
c^2 = 169
Pero eso no puede ser la respuesta porque no tiene sentido a que x sea un lado que mide 169 entonces tenemos que usar sqrt rt
sqrt rt de 169 = 13
x = 13
I would help but I’m not sure how, please do not press that link or else it will track your address! <33
Answer:
3x+s
should be the answer hope it helps
Answer:
x = 16 2/3
Step-by-step explanation:
The two triangles in the picture are <em>similar</em>, which means that the corresponding side lengths on each of them are proportional - the ratios between them are equal to each other.
On the larger triangle, let's pick the pair of lengths 25 and 39. The corresponding pair on the triangle has the lengths x and 26. Similarity then tells us that the ratios, 25/39 and x/26, are equivalent. All that's left to do now is set up an equation and solve for x:
![\frac{25}{39}=\frac{x}{26}](https://tex.z-dn.net/?f=%5Cfrac%7B25%7D%7B39%7D%3D%5Cfrac%7Bx%7D%7B26%7D)
multiplying both sides of the equation by 26 gives us
![26\left(\frac{25}{39}\right)=x](https://tex.z-dn.net/?f=26%5Cleft%28%5Cfrac%7B25%7D%7B39%7D%5Cright%29%3Dx)
We can do a little reduction here: 26 and 39 have a common factor of 13, and since 26 = 13 x 2 and 39 = 13 x 3, the expression on the left reduces to
![2\left(\frac{25}{3}\right)=x](https://tex.z-dn.net/?f=2%5Cleft%28%5Cfrac%7B25%7D%7B3%7D%5Cright%29%3Dx)
We can rewrite 25/3 as the mixed number 8 1/3, making our equation
![2\left(8\frac{1}{3}\right) =x](https://tex.z-dn.net/?f=2%5Cleft%288%5Cfrac%7B1%7D%7B3%7D%5Cright%29%20%3Dx)
So finally, we have ![x=16\frac{2}{3}](https://tex.z-dn.net/?f=x%3D16%5Cfrac%7B2%7D%7B3%7D)