F(x) = x - 7 && h(x) = 2x + 3
h(f (x)) = 2(f(x)) + 3
h(f (x)) = 2(x - 7) + 3
h(f (x)) = 2x - 14 + 3
h(f (x)) = 2x - 11
The answer is A
Answer:
x²-10x+25
Step-by-step explanation:
f(x) = x²
<h3>i) substitute the x-5 into x</h3>
f(x-5) = (x-5)²
<h3>ii) using the formula (a-b)² = a²-2ab+b², expand the expression</h3>
f(x-5) = (x-5)²
= x²-2(5x)+5²
<h3>iii) calculate the products</h3>
f(x-5) = x²-2(5x)+5²
= x²-10x+25
* the pic i inserted shows the shape of your graph
That would be 263 idk I just had to put an answer so that it lets me see more answers
Answer:
y(t) = 2.5 e⁶ᵗ + 2.5 e⁻⁶ᵗ
Or
y(t) = 5 e⁻⁶ᵗ
Step-by-step explanation:
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ
Let us find our value for y(t) that satisfies the conditions
1) y" - 36y = 0
y" = (d²y/dt²)
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ
y' = (dy/dt) = 6c₁ e⁶ᵗ - 6c₂ e⁻⁶ᵗ
y" = (d/dt)(dy/dt) = 36c₁ e⁶ᵗ + 36c₂ e⁻⁶ᵗ
y" - 36y = 36c₁ e⁶ᵗ + 36c₂ e⁻⁶ᵗ - 36(c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ) = 36c₁ e⁶ᵗ + 36c₂ e⁻⁶ᵗ - 36c₁ e⁶ᵗ - 36c₂ e⁻⁶ᵗ = 0.
The function satisfies this condition.
2) y(0) = 5
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ
At t = 0
y(0) = c₁ e⁰ + c₂ e⁰ = 5
c₁ + c₂ = 5 (e⁰ = 1)
3) lim t→+[infinity] y(t)=0
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ = 0 as t→+[infinity]
c₁ e⁶ᵗ = - c₂ e⁻⁶ᵗ as t→+[infinity]
c₁ = - c₂ e⁻¹²ᵗ as t→+[infinity]
e⁻¹²ᵗ = 0 as t→+[infinity]
c₁ = c₂ or c₁ = 0
Recall c₁ + c₂ = 5
If c₁ = 0, c₂ = 5
If c₁ = c₂, c₁ = c₂ = 2.5
y(t) = c₁ e⁶ᵗ + c₂ e⁻⁶ᵗ = 2.5 e⁶ᵗ + 2.5 e⁻⁶ᵗ
Or
y(t) = 5 e⁻⁶ᵗ
Step-by-step explanation:
x² + 2x - 5 = 0
that is the same as
x² + 2x + 1 = 6
(x + 1)² = 6
x + 1 = ±sqrt(6)
x = -1 ± sqrt(6)
x1 = -1 + sqrt(6)
x2 = -1 - sqrt(6)
n² - 20n + 2 = 4
this is the same as
(n - 10)² = 102
n - 10 = ±sqrt(102)
n = 10 ± sqrt(102)
n1 = 10 + sqrt(102)
n2 = 10 - sqrt(102)