Write a number as prime factors means to write the number as a product of numbers, all of which are prime. We start by checking whether the number is divisible by prime numbers, starting from the smallest prime number,2.
let's divide 24 into its factors.
first, it's even, so it must divide by 2
24=2*12
12 is also even, so it must divide by 2:
24=2*12=2*2*6
6 is also even, so it must divide by 2:
24=2*12=2*2*6=2*2*2*3
3 is not even, but it's a prime number.
so the solution is
2*2*2*3
Answer:
The correct option is A.
Step-by-step explanation:
Domain:
The expression in the denominator is x^2-2x-3
x² - 2x-3 ≠0
-3 = +1 -4
(x²-2x+1)-4 ≠0
(x²-2x+1)=(x-1)²
(x-1)² - (2)² ≠0
∴a²-b² =(a-b)(a+b)
(x-1-2)(x-1+2) ≠0
(x-3)(x+1) ≠0
x≠3 for all x≠ -1
So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong
Asymptote:
x-3/x^2-2x-3
We know that denominator is equal to (x-3)(x+1)
x-3/(x-3)(x+1)
x-3 will be cancelled out by x-3
1/x+1
We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....
Step-by-step explanation:
this is all I can find I hope it helps you out us not I'm sorry
Evaluate 0.3 y+\dfrac yz0.3y+ z y 0, point, 3, y, plus, start fraction, y, divided by, z, end fraction when y=10y=10y, equals,
sergij07 [2.7K]
Answer:
5
Step-by-step explanation:
Put the numbers where the corresponding variables are in the formula and do the arithmetic.
The expression evaluates to 5.
5.99999999999999999 is close to 6
