Answer:
9:15
Step-by-step explanation:
![\frac{3*3}{5*3}=\frac{9}{15}=9:15](https://tex.z-dn.net/?f=%5Cfrac%7B3%2A3%7D%7B5%2A3%7D%3D%5Cfrac%7B9%7D%7B15%7D%3D9%3A15)
If
is the first number in the progression, and
is the common ratio between consecutive terms, then the first four terms in the progression are
![\{x,xr,xr^2,xr^3\}](https://tex.z-dn.net/?f=%5C%7Bx%2Cxr%2Cxr%5E2%2Cxr%5E3%5C%7D)
We want to have
![\begin{cases}xr^2-x=12\\xr^3-xr=36\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dxr%5E2-x%3D12%5C%5Cxr%5E3-xr%3D36%5Cend%7Bcases%7D)
In the second equation, we have
![xr^3-xr=xr(r^2-1)=36](https://tex.z-dn.net/?f=xr%5E3-xr%3Dxr%28r%5E2-1%29%3D36)
and in the first, we have
![xr^2-x=x(r^2-1)=12](https://tex.z-dn.net/?f=xr%5E2-x%3Dx%28r%5E2-1%29%3D12)
Substituting this into the second equation, we find
![xr(r^2-1)=12r=36\implies r=3](https://tex.z-dn.net/?f=xr%28r%5E2-1%29%3D12r%3D36%5Cimplies%20r%3D3)
So now we have
![\begin{cases}9x-x=12\\27x-3x=36\end{cases}\implies x=\dfrac32](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D9x-x%3D12%5C%5C27x-3x%3D36%5Cend%7Bcases%7D%5Cimplies%20x%3D%5Cdfrac32)
Then the four numbers are
![\left\{\dfrac32,\dfrac92,\dfrac{27}2,\dfrac{81}2\right\}](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cdfrac32%2C%5Cdfrac92%2C%5Cdfrac%7B27%7D2%2C%5Cdfrac%7B81%7D2%5Cright%5C%7D)
Answer:
Reflexive property of congruence.
Step-by-step explanation:
Answer:
Step-by-step explanation:
√a+√b=√(a+b)
squaring
a+b+2√a√b=a+b
or 2√a√b=0
which is possible only if a=0 or b=0
Both a and b can't be zero as a≠b