Please Help! My question is in the picture.

Mathematics

1 answer:

Varvara68 [4.7K]3 years ago

4 0

First we need to determine the type of progression in the question.
That's geometric progression. Because the pattern from one sequence to the others are about multiplying.

Second, determine the ratio of the progression
r = a₂/a₁
r = a₂ ÷ a₁
r = 1/2 ÷ 2
r = 1/2 × 1/2
r = 1/4

Third, determine the formula to know the recursive rule
a₂ = a × 1/4
a₂ = 1/4 × a
$a_{n} = \frac{1}{4} a_{(n-1)}$

Fourth, determine a₁. a₁ is the first term of the progression
a₁ = 2

Final answer:
Recursive rule
$a_{n} = \frac{1}{4} a_{(n-1)}$
a₁ = 2

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$y' = \frac{dy}{dx}$

seperable differential equations will have the form
$\frac{dy}{dx} = F(x) G(y)$

what you do from here is isolate all the y terms on one side and all the X terms on the other
$\frac{dy}{G(y)} = F(x) dx$
just divided G(y) to both sides and multiply dx to both sides

then integrate both sides
$\int \frac{1}{G(y)} dy = \int F(x) dx

$

once you integrate, you will have a constant. use the initial value condition to solve for the constant, then try to isolate x or y if the question asks for it

In your problem,
$G(y) = \sqrt{1-y^2}

F(x) = 2x$

so all you need to integrate is
$\int \frac{1}{\sqrt{1-y^2}} dy = \int 2x dx$