Volume of ball = 13.39 inch ^3 Volume of ball = 4/3 X pi x r^3 ATQ 4/3 X pi X r^3 = 13.39 4/3 X 22/7 X r^3 = 13.39 r^3 = 13.39 x 7/22 x 3/4 r^3 = 281.19/88 r^3 = 3.19 r = 1.47 inches d = 2 x 1.47 = 2.94 inches
Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]
⇒ a + (9-1)d = 0
⇒ a + 8d = 0 ⇒ a = -8d ...(i)
Now, its 19th term , T19 = a + (19-1)d
= - 8d + 18d [from Eq.(i)]
= 10d ...(ii)
and its 29th term, T29 = a+(29-1)d
= -8d + 28d [from Eq.(i)]
= 20d = 2 × T19
Hence, its 29th term is twice its 19th term
12 cm to 24 cm
the scale factor is 2x
Step-by-step explanation:
a) determine the rate in km/L
==> 9.5 L/100km = 10.526316 km/L
b) how far can the car travel with 475L petrol
==> 0.210526315789473km/l
c) how many litres of petrol do they need to travel 350 km
==>Liters per km (l/km)2.86×10-3
Liters per 10 km (l/10 km)0.03
Liters per 100 km (l/100km)0.29
Kilometer per liter (km/l)350
