Start with 180.
<span>Is 180 divisible by 2? Yes, so write "2" as one of the prime factors, and then work with the quotient, 90. </span>
<span>Is 90 divisible by 2? Yes, so write "2" (again) as another prime factor, then work with the quotient, 45. </span>
<span>Is 45 divisible by 2? No, so try a bigger divisor. </span>
<span>Is 45 divisible by 3? Yes, so write "3" as a prime factor, then work with the quotient, 15 </span>
<span>Is 15 divisible by 3? [Note: no need to revert to "2", because we've already divided out all the 2's] Yes, so write "3" (again) as a prime factor, then work with the quotient, 5. </span>
<span>Is 5 divisible by 3? No, so try a bigger divisor. </span>
Is 5 divisible by 4? No, so try a bigger divisor (actually, we know it can't be divisible by 4 becase it's not divisible by 2)
<span>Is 5 divisible by 5? Yes, so write "5" as a prime factor, then work with the quotient, 1 </span>
<span>Once you end up with a quotient of "1" you're done. </span>
<span>In this case, you should have written down, "2 * 2 * 3 * 3 * 5"</span>
X=10 is the answer to this
1. There is no smallest number
2. 0 means that there is nothing. It's a neutral real number
3. 2
4. -2
5. Fractions may or may not be integers
6. 1/3, 4/5, 7/16, 16/21
Step-by-step explanation:
y = |x - 2| - 1
Given equation is in the form of y=|x-h|+k
Where (h,k) is the vertex
h = 2 and k =-1
So (2, -1) is the vertex for the given function
To find x intercepts we plug in 0 for y
we will get two values for x because of absolute function
0 = |x-2| -1
add 1 on both sides
|x-2| =1
x-2 =1 and x-2 =-1
x= 3 and x= 1
So x intercepts are (1,0) and (3,0)
the graph is attached below
