The answer to the question is 12,400
Answer:
I assume you know Arithmetic Progression .
so, we have to find the first and last 4-digit number divisible by 5
first = 1000 , last = 9990
we have a formula,
= a + (n-1)d
here,
is the last 4-digit number divisible by 5.
n is the number of 4-digit even numbers divisible by 5
d is the common difference between the numbers, which is 10 in this case
a is the first 4-digit number divisible by 5
9990 = 1000 + (n-1)*10
899 = n-1
n = 900
Hence, there are 900 4-digit even numbers divisible by 5
Answer:
p = ½ (x₁ + x₂)
q = a (x₁x₂ − ¼ (x₁ + x₂)²)
Step-by-step explanation:
y = a (x − x₁) (x − x₂)
Expand:
y = a (x² − x₁x − x₂x + x₁x₂)
y = a (x² − (x₁ + x₂)x + x₁x₂)
Distribute a to the first two terms:
y = a (x² − (x₁ + x₂)x) + ax₁x₂
Complete the square:
y = a (x² − (x₁ + x₂)x + ¼(x₁ + x₂)²) + ax₁x₂ − ¼ a(x₁ + x₂)²
y = a (x − ½ (x₁ + x₂))² + a (x₁x₂ − ¼ (x₁ + x₂)²)
Therefore:
p = ½ (x₁ + x₂)
q = a (x₁x₂ − ¼ (x₁ + x₂)²)
Answer:
Choice A
Step-by-step explanation:
(4m^5n^2/m^2n)^3
dividing exponents = subtraction
(4m^3n)^3
4^3 = 64
(m^3 )^3 = m^3x3 = m^9
64m^9n^3