Answer:
C
Step-by-step explanation:
There is a common ratio between consecutive term in the sequence, that is
÷
= 3 ÷
= 30 ÷ 3 = 300 ÷ 30 = 10
This indicates the sequence is geometric with n th term
= a
where a is the first term and r the common ratio
Here a = and r = 10 , thus
=
= ×
= 3 × ×
= 3 ×
Thus
= 3
→ C
Answer:
Question 9
(a) The distance, Jalaj walks in one day is 4.4 km
(b) The amount Jalaj raises after walking for 22 km at the end of the 5 days is $8
Question 10
(b) The difference between the largest and smallest areas is 2,400 m²
Step-by-step explanation:
Question 9
(a) The distance Jalaj walks in 5 days = 22 km
Whereby Jalaj walks equal distance every day, we have;
The distance, Jalaj walks in one day = 22 km/5 days = 4.4 km/day
The distance, Jalaj walks in one day = 4.4 km
(b) The amount he raises for every kilometer he walks = $1.60
The amount he raises after walking for 22 km at the end of the 5 days = 5 × $1.60 = $8
Question 10
(b) The given side length of the square = 120 meters to the nearest 10 meters
Therefore;
The maximum dimension for the side length of the square = 120 + 10/2 = 125
The largest possible area of the square, = 125 m × 125 m = 15,625 m²
The minimum dimension for the side length of the square = 120 m - 10 m/2 = 115 m
The smallest possible area of the square, = 115 m × 115 m = 13,225 m².
The difference between the largest and smallest areas, -
= 15,625 - 13,225 = 2,400 m².
Answer:
Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box has largest volume.
Step-by-step explanation:
Given that,
A box is being created out of a 15 inches by 10 inches sheet of metal.
The length of the one side of the squares which are cut out of the each corners of the metal sheet be x.
The length of the metal box be = (15-2x) inches.
The width of the metal box be =(10-2x) inches
The height of the metal box be =x inches
Then, the volume of the metal box= length×width×height
=(15-2x)(10-2x)x cubic inches
=(150x-50x²+4x³) cubic inches
∴ V= 4x³-50x²+15x
Differentiating with respect to x
V'=12x²-100x+15
Again differentiating with respect to x
V''=24x-100
For maximum or minimum value, V'=0
12x²-100x+15=0
Apply quadratic formula , here a=12, b= -100 and c=15
For x= 8.18, The value of (15-2x) and (10-2x) will negative.
∴x=0.1528 .
Now,
∴At x=0.1528 inch, the volume of the metal box will be maximum.
Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box has largest volume.