equation is g(x)=4(x-2)+7, (bit unsure of this next part) rule is g(x-2)
Answer:
a = l²
v = s³
Step-by-step explanation:
The area of a rectangle is the product of its length and width. When that rectangle is a square, the length and width are the same. Here, they are given as "l". Then the area of the square is ...
a = l·l = l²
__
The volume of a cuboid is the product of its height and the area of its base. A cube of edge length s has a square base of side length s and a height of s. Then its volume will be ...
v = s·(s²) = s³
The two equations you want are ...
• a = l²
• v = s³
<span>f(x) = x</span>² <span>+ 12x + 6 </span>→ y = x² + 12x + 6<span>
Let us convert the standard form into vertex form.
1) Complete the squares. Isolate x</span>² and x terms.
<span>y - 6 = x</span>² + 12x
<span>
2) Create the perfect square trinomial. Whatever number is added on one side must also be added on the other side.
y - 6 + 36 = x</span>² + 12x + 36<span>
y + 30 = (x + 6)</span>²
<span>y = (x + 6)</span>² - 30 ← Vertex form
<span>
To check:
y = (x + 6) (x + 6) - 30
y = x</span>² + 6x + 6x + 36 - 30
<span>y = x</span>² + 12x + 6<span>
The zero that could be added to the given function is 36, -36</span>
Answer:
the ordered pairs should be y and x
Step-by-step explanation:
Answer:
- 18x + 15y ≤ 300
- x ≥ 0; y ≥ 6
Step-by-step explanation:
Variables are defined in the problem statement.
18x +15y ≤ 300 . . . . total budget
y ≥ 6 . . . . . . . . . . . . minimum number of annuals
x ≥ 0 . . . . . number of perennials cannot be negative
This system of inequalities describes the situation.
