We know that the formula for calculating distance is:
distance d = velocity v * time t
Calculating for each interval:
1: 25 km/hr for 4 minutes
d1 = (25 km / 60 min) (4 min) = 1.67 km
2: 50 km/hr for 8 minutes
d2 = (50 km / 60 min) (8 min) = 6.67 km
3: 20 km/hr for 2 minutes
d2 = (20 km / 60 min) (2 min) = 0.67 km
The total distance covered is:
d = d1 + d2 + d3
d = 1.67 + 6.67 + 0.67
d = 9 km
The average speed is:
s = 9 km (1000 m / km) / (14 min * 60 s / min)
s = 10.72 m / s
The answer is A for the correct answer
X/15 = 10/20
20x = 150
x = 150/20
x = 7.5
Answer: Multiply the divisor and dividend by 10
Answer:
Below.
Step-by-step explanation:
f) (a + b)^3 - 4(a + b)^2
The (a+ b)^2 can be taken out to give:
= (a + b)^2(a + b - 4)
= (a + b)(a + b)(a + b - 4).
g) 3x(x - y) - 6(-x + y)
= 3x( x - y) + 6(x - y)
= (3x + 6)(x - y)
= 3(x + 2)(x - y).
h) (6a - 5b)(c - d) + (3a + 4b)(d - c)
= (6a - 5b)(c - d) + (-3a - 4b)(c - d)
= -(c - d)(6a - 5b)(3a + 4b).
i) -3d(-9a - 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b).
= (3d + 2c)(9a + 2b).
j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)
= a^2b^3(2a + 1) + 6ab^2(2a + 1)
= (2a + 1)( a^2b^3 + 6ab^2)
The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:
(2a + 1)ab^2(ab + 6)
= ab^2(ab + 6)(2a + 1).
