a. The moduli are coprime, so you can apply the Chinese remainder theorem directly. Let
- Taken mod 3, the last two terms vanish, and
so we need to multiply by the inverse of 2 modulo 3 to end up with a remainder of 1. Since 
, we multiply the first term by 2.
- Taken mod 4, the first and last terms vanish, and
. Multiply by the inverse of 3 modulo 4 (which is 3 because 
), then by 2 to ensure the proper remainder is left.
- Taken mod 5, the first two terms vanish, and
. Multiply by the inverse of 2 modulo 5 (3, since 
) and again by 3.
By the CRT, we have
i.e. any number 
(where 
is an integer) satisifes the system.
b. The moduli are not coprime, so we need to check for possible contradictions. If 
and 
, then we need to have 
. This basically amounts to checking that if , then we should also have 
.
The last congruence conflicts with the previous one modulo 3, so there is no solution to this system.
She attempted 120 free throws.
Answer:
The solutions to the system of equations are:
Step-by-step explanation:
Given the system of the equations
isolate 'x' for 2x-y
Add y to both sides
Divide both sides by 2
isolate 'y' for 
subtract y from both sides
Divide both sides by 9
The solutions to the system of equations are:
Answer:
Step-by-step explanation:
![\displaystyle \frac{3\sqrt{3}}{2}a^2 = A \hookrightarrow \frac{3\sqrt{3}}{2}6^2 = A \\ \frac{3\sqrt{3}}{2}[36] = A; 54\sqrt{3}\:[or\:93,530743609...] \\ \\ \boxed{93,5 \approx A}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7Da%5E2%20%3D%20A%20%5Chookrightarrow%20%5Cfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D6%5E2%20%3D%20A%20%5C%5C%20%5Cfrac%7B3%5Csqrt%7B3%7D%7D%7B2%7D%5B36%5D%20%3D%20A%3B%2054%5Csqrt%7B3%7D%5C%3A%5Bor%5C%3A93%2C530743609...%5D%20%5C%5C%20%5C%5C%20%5Cboxed%7B93%2C5%20%5Capprox%20A%7D)
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