Range is ouput
testing limints
basically see the vertices
y=a(x-h)^2+k
(h,k) is vertex
also, when a is negative, graph opens down
when a is positive, graph opens up
y<u><</u>5
k musbe 5
eliminate last 2
since less than, must open down
negative a
second one
f(x)=-(x-4)^2+5 is ansre
Answer:
6.5 square units
Step-by-step explanation:
There are MANY ways to do this problem, most of them involving breaking it into 3 or four shapes. But this one is easier if you do most of it by taking out the squares and half squares.
Squares - 3
Half squares - 4 (totals 2 squares)
Triangle - 1 - 1/2bh = 1/2x1x3 = 1.5
TOTAL = 3+2+1.5=6.5
Answer:
Hello so.
65.4 x 2.3 = 150.42
Step-by-step explanation:
The Images I have attached show the work.
Answer:
Step-by-step explanation:
Discount
Sale price
To find the equation of a line that is parallel to your original equation and goes through a certain point on a graph, here's what you need to know:
First you need to find the slope of your original equation.
To do that, you need to convert it to slope intercept form (y = mx+b).
Add the x over, and then divide everything by 5 to get the y by itself.
Here's what that would look like (without the small steps that I mentioned):
-x + 5y = 25
5y = x + 25
y = 1/5x + 5
That's the original equation rewritten in slope intercept form.
The m represents the slope, so this equation's slope is 1/5.
Because you are given a point, and now you have a slope, the best and easiest route is using point slope form.
I've seen different versions of the equation base but I prefer y - y(sub1) = m(x - x(sub1))
But since I can't use subscripts in this, I'll use the one with h and k. The h is the x value of the point, and the k is the y value.
(h,k)
Then just substitute the values in and solve for y.
y - k = m(x - h)
y + 5 = 1/5(x + 5)
y + 5 = 1/5x + 1
y = 1/5x - 4
Your final answer is
y = 1/5x - 4
You can double check by using a graph. If the slopes are the same, the lines should be parallel.
I hope that helps. If anything didn't make sense, feel free to ask me.