You start with the basic y=x^2.
That’s the parabola with a vertex at (0,0), opening up, etc.
The transformation aspect is the “+7” portion. This “+7” shifts the entire graph of y=x^2 up by 7 units.
The vertex is now (0,7), it still opens up, etc.
It’s kind of a silly question for your teacher to ask when the graph given only goes up to 6.
First, let's establish a ratio between these two values. We'll use that as a starting point. I personally find it easiest to work with ratios as fractions, so we'll set that up:

To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:

We have our answer: approximately
0.14 cm or
1.4 mm per year
Answer:
vector VW= w - v
vector WV= v - w
Step-by-step explanation:
there are no figures to solve the question so i'll just write the formula..
vector VW= w - v
vector WV= v - w
Answer:
x = 14
Step-by-step explanation:
given f(x) = 2(3x - 5) and f(x) = 74 , then equating gives
2(3x - 5) = 74 ( divide both sides by 2 )
3x - 5 = 37 ( add 5 to both sides )
3x = 42 ( divide both sides by 3 )
x = 14
I think it should be x=12.
The equation should be 6x-2+20°=90°