She went on 8 rides and 5 games because 2.25*8=18 1.25*5=6.25
18.00+6.25=24.25 and 8-5=3.
Answer:
Step-by-step explanation:
|4x-3|=5√(x+4) ⇔ |4x-3|²=5²(√(x+4))² and x+4 ≥ 0
⇔ (4x-3)² = 25(x+4) and x+4 ≥ 0 ( because : /a/² = a²)
⇔16x²-24x+9 = 25x +100 and x+4 ≥ 0
⇔ 16x² -49x - 91 =0 and x+4 ≥ 0 quadratic equation
Δ = (-49)²-4(16)(-91) = 8225
two solution : X1 = (49-√8225)/32 ≅ - 1.3 accept (-1.3+4 ≥ 0)
X2 = (49+√8225)/32 ≅4.37 accept (4.37+4 ≥ 0)
Answer: Dimensions of A are of length [L]
Dimensions of B are of 
Dimensions of C are of 
Step-by-step explanation:
The given equation is
Since the dimension on the L.H.S of the equation is [L] , each of the terms on the right hand side should also have dimension of length[L] to be dimensionally valid
Thus
Dimensions of A = [L]
Dimensions of Bt = [L]
![Bt=[L]\\\\](https://tex.z-dn.net/?f=Bt%3D%5BL%5D%5C%5C%5C%5C)
![[B][T]=[L]](https://tex.z-dn.net/?f=%5BB%5D%5BT%5D%3D%5BL%5D)
![\\\\\therefore [B]=LT^{-1}](https://tex.z-dn.net/?f=%5C%5C%5C%5C%5Ctherefore%20%5BB%5D%3DLT%5E%7B-1%7D)
Similarly
Dimensions of ![Ct^{}2 = [L]](https://tex.z-dn.net/?f=Ct%5E%7B%7D2%20%3D%20%5BL%5D)
![Ct^{2}=[L]\\\\[C][T]^{2}=[L]\\\\\therefore [C]=LT^{-2}](https://tex.z-dn.net/?f=Ct%5E%7B2%7D%3D%5BL%5D%5C%5C%5C%5C%5BC%5D%5BT%5D%5E%7B2%7D%3D%5BL%5D%5C%5C%5C%5C%5Ctherefore%20%5BC%5D%3DLT%5E%7B-2%7D)
Answer:
24 units
Step-by-step explanation:
Original perimeter:
5+4+3=12
When each side length is doubled, the original perimeter is doubled.
12*2=24
Answer : 2ab^3+4a^2b-5ab+5
