She read 5.5 books per month x
9514 1404 393
Answer:
1 1/5
Step-by-step explanation:
We want to find the number of times (5/6) goes into 1, so we divide ...
1/(5/6) = 1 × 6/5 = 6/5 = 1 1/5
There are 1 1/5 groups of 5/6 in 1.
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<em>Another way to look at it</em>
You can consider 1 to be divided into 6 equal part:
o o o o o o
We can group these in groups of 5, so each group is 5/6 of the whole:
(o o o o o) o
What we see is that we have 1 group, and 1/5 of another group.
That is, 1 has 1 1/5 groups of 5/6 in it.
To solve this problem you need to know the law of cosines.
Answer:
![y {}^{6}](https://tex.z-dn.net/?f=y%20%7B%7D%5E%7B6%7D%20)
Because you multiplied two same things. So it removes the square root and then just add the coefficients.
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