Answer:
x > -8
Step-by-step explanation:
5 - (x+5) > -2(x+4)
5 - x - 5 > -2x -8
- x > -2x -8
-x +2x > -8
x > -8
Answer:
a^2 -25
Step-by-step explanation:
5a-25 + a^2 - 5a
a^2 - 25
0.373 seconds. First, calculate the initial vertical velocity of the shell. 800sin(30) = 800*0.5 = 400 m/s Now the formula for the distance traveled is d = 400 m/s * T + 0.5A T^2 Substituting known values gives. 150 = 400 m/s * T + 0.5*9.80m/s^2 T^2 150 = 400 m/s * T + 4.9 m/s^2 T^2 Arrange as a quadratic formula 0 = 400 m/s * T + 4.9 m/s^2 T^2 - 150 4.9 m/s^2 T^2 + 400 m/s * T - 150 = 0 Now solve for T using the quadratic formula with a=4.9, b=400, and c=-150 The calculated value is 0.373 seconds. Is this value reasonable? Let's check. The initial downward velocity is 400 m/s. So 150/400 = 0.375 seconds. Since the actual time will be a bit less due to acceleration by gravity and since the total time is so short, there won't be much acceleration due to gravity, the value of 0.373 is quite reasonable.
Answer:
10 customers
Step-by-step explanation:
16x+4y=360
x=20
16(20)=320
320+4y=360
4y=40
y=10
1. Area of the first figure;
Area of the quarter circle = 1/4 πr²
= 1/4π 12²
= 36π cm²
Area of the triangle = 1/2×base×height
= 1/2 × 12×12
= 72 cm²
Thus the area of the region
= (36π-72) cm²
2. Perimeter of the first figure
The length of the arc AC is 2πr/4
= 2π×12/4
= 6π cm
the length of line AC
√(12² +12²) = 16.97
Thus the perimeter of the figure
= (6π+16.97) cm
3. The are of figure 2
Area of the semi circle πr²/2
= 12²π/2 = 72π cm²
The area of the square
= 24 × 24 =576 cm²
Area of the semi circle = 72 π cm²
Thus the area of the part of the square
= (576-72π) cm²
total area
72π +576-72π
= 576 cm²
4. Perimeter of figure 2
The length of arc AD = (2πr)/2
= 2π×12/2 = 12π cm
but length of arc AD is equal to that of arc BC
Thus arc BC = 12πcm
Therefore, total perimeter
= 12π +12π +24 +24
= (24π +48) cm