Answer:
70 cars per year.
Step-by-step explanation:
Given:
In 1984 a locally-owned car company sold 2,743 cars.
In 2003, the car sales rose to 4,069.
Question asked:
What was the average rate of change for the total number of cars sold?
Solution:
In 2003 number of cars sold = 4069
In 1984 number of cars sold = 2743
Change in the number of car sold = 4069 - 2743
= 1326
Now, time interval between 2003 and 1984 = 19
Thus, the average rate of change for the total number of cars sold is 70 cars per year.
Answer:
10 times greater is the value of 5 in 3,590 than the value of 5 in 359.
Step-by-step explanation:
To find : How many times greater is the value of 5 in 3,590 than the value of 5 in 359 ?
Solution :
The place value stem is
Thousand Hundred Tens Ones
1000 100 10 1
The value of 5 in 3,590 is at hundred place
So, 
The value of 5 in 359 is at tens place
So, 
Number of time greater is the value of 5 in 3,590 than the value of 5 in 359 is given by,
Therefore, 10 times greater is the value of 5 in 3,590 than the value of 5 in 359.
Answer:
d = 
Step-by-step explanation:
Given that W varies jointly as L and d² then the equation relating them is
W = kLd² ← k is the constant of variation
To find k use the condition W = 140 when d = 4 and L = 54, thus
140 = k × 54 × 4² = 864k ( divide both sides by 864 )
= k , that is
k = 
W = Ld² ← equation of variation
Multiply both sides by 216
216W = 35Ld² ( divide both sides by 35L )
= d² ( take the square root of both sides )
d =
Answer:
B. 3.6 x 10^3
Step-by-step explanation:
(1.2 x 10^-2) x (3 x 10^5)
When we multiply terms with powers , we multiply the factors out front and add the exponents
a * 10^b * c* 10^d = ac * 10^(b+d)
(1.2 x 10^-2) x (3 x 10^5) = (1.2* 3) * 10^(-2+5)
= 3.6 * 10 ^3
