Answer:
What happens when h is positive?
The asymptote of the graph shifts <em>right</em> by that amount.
What happens when h is negative?
The asymptote of the graph shifts <em>left</em> by that amount.
What happens when k is positive?
The entire graph shifts <em>up</em> by that amount.
What happens when k is negative?
The entire graph shifts <em>down</em> by that amount.
First, find what you can multiply by 
to get closest to (but not over) 
.
This value is 
.
Now, multiply 
to get 
.
Subtract 
to get a remainder of 
.
15 and the error is on step 3
Reasoning:
I’m not certain how they got 1.5 but you basically divide both sides by two, the 2 would be canceled and you would be left with w, divide 30 by 2 and you get w=15, sorry i suck at explaining
Answer:
remaining length: 5y² + 6y + 12
Explanation:
new length = original length - length cut
solve for new length:
6y² + 6y + 1 - (y² - 11)
6y² + 6y + 1 - y² + 11
6y² - y² + 6y + 11 + 1
5y² + 6y + 12
Hello,
Let's suppose y'=y-3
and x'=x-5
Parabola become y'²=8x' whose focus is (8/4,0)=(2,0)
y=y'+3==>y=0+3=3
x=x'+5==>x=2+5=7
Focus is (7,3)
Rem for parabola y²=kx focus is (k/4,0)
