11
(9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99)
Those numbers divisible by 9 are the multiple of 9; thus need to know how many multiples of 9 there are between 1 and 100:
100 ÷ 9 = 11 r 1 ÷ last multiple of 9 is 11 × 9 (= 99)
→ There are 11 - 1 + 1 = 11 numbers between 1 and 100 which are divisible by 9.
(They are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.)
Answer:
5. 8
6. 18
Step-by-step explanation:
5.
a(a^2 + a - 1) + 7 =
= a^3 + a^2 - a + 7
For a = -1,
= (-1)^3 + (-1)^2 - (-1) + 7
= -1 + 1 + 1 + 7
= 8
6.
2ab(a + b) - 3ab(a - b) =
= 2a^2b + 2ab^2 - 3a^2b + 3ab^2
= -a^2b + 5ab^2
For a = 1 and b = 2,
= -(1)^2(2) + 5(1)(2)^2
= -1(2) + 5(4)
= - 2 + 20
= 18
Answer:
100 or 150 units of the product.
Step-by-step explanation:
R = x(1000-4x)
What is the value of x if R = 60,000?
x(1000-4x) = 60000
-4x^2 + 1000x = 60000
x^2 - 250x = -15000
x^2 - 250x + 15000 = 0
Factorize:
(x-100)(x-150) = 0
x-100 = 0 V x-150 = 0
x = 100 V x = 150
Answer:
i think its -1
Step-by-step explanation:
sorry if im wrong
Answer:
HCF of 60 and 80 is 20.
Step-by-step explanation:
