Note: The equations written in this questions are not appropriately expressed, however, i will work with hypothetical equations that will enable you to solve any problems of this kind.
Answer:
For the system of equations to be unique, s can take all values except 2 and -2
Step-by-step explanation:

![\left[\begin{array}{ccc}2s&4\\2&s\end{array}\right] \left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right] = \left[\begin{array}{ccc}-3 \\6 \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2s%264%5C%5C2%26s%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%20%5C%5Cx_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%20%5C%5C6%20%5Cend%7Barray%7D%5Cright%5D)
For the system to have a unique solution, 

Answer:
The fourth one
Step-by-step explanation:
Only the first one x and y interval 1, 2. The rest aren't linear.
Answer:
3/10???????????????????????
Answer:
V=25088π vu
Step-by-step explanation:
Because the curves are a function of "y" it is decided to take the axis of rotation as y
, according to the graph 1 the cutoff points of f(y)₁ and f(y)₂ are ±2
f(y)₁ = 7y²-28; f(y)₂=28-7y²
y=0; x=28-0 ⇒ x=28
x=0; 0 = 7y²-28 ⇒ 7y²=28 ⇒ y²= 28/7 =4 ⇒ y=√4 =±2
Knowing that the volume of a solid of revolution V=πR²h, where R²=(r₁-r₂) and h=dy then:
dV=π(7y²-28-(28-7y²))²dy ⇒dV=π(7y²-28-28+7y²)²dy = 4π(7y²-28)²dy
dV=4π(49y⁴-392y²+784)dy integrating on both sides
∫dV=4π∫(49y⁴-392y²+784)dy ⇒ solving ∫(49y⁴-392y²+784)dy
49∫y⁴dy-392∫y²dy+784∫dy =
V=4π(
) evaluated -2≤y≤2, or 2(0≤y≤2), also
⇒ V=25088π vu