7. y = 2x² + 3 function not function ?

If the graph of y=f(x) passes through the point (0,1), and dy/dx=xsin(x^2)/y, then f(x)= ?

Zigmanuir [339]

The differential equation

dy/dx = x sin(x ²) / y

is separable as

y dy = x sin(x ²) dx

Integrate both sides:

∫ y dy = ∫ x sin(x ²) dx

∫ y dy = 1/2 ∫ 2x sin(x ²) dx

∫ y dy = 1/2 ∫ sin(x ²) d(x ²)

1/2 y ² = -1/2 cos(x ²) + C

Solve for y implicitly:

y ² = -cos(x ²) + C

Given that y = 1 when x = 0, we get

1² = -cos(0²) + C

1 = -cos(0) + C

1 = -1 + C

C = 2

Then the particular solution to the DE is

y ² = 2 - cos(x ²)

Solving explicitly for y would give two solutions,

y = ± √(2 - cos(x ²))

but only the one with the positive square root satisfies the initial condition:

√(2 - cos(0²)) = √1 = 1

-√(2 - cos(0²)) = -√1 = -1 ≠ 1

So the unique solution to this DE is

y = √(2 - cos(x ²))

4 0

3 years ago

Which is the graph of an odd monomial function?

Minchanka [31]

Odd function must pass through origin so first two options are not correct.

Odd function must be rotational symmetry around origin. So last option is not correct.

So answer is C.

6 0

3 years ago

Mohamed decided to track the number of leaves on the tree in his backyard each year The first year there were 500 leaves Each ye

svetlana [45]

Answer:

The required recursive formula is

$f(n)= 500\times(1.4)^{n-1}\\$

Step-by-step explanation:

Mohamed decided to track the number of leaves on the tree in his backyard each year.

The first year there were 500 leaves

$Year \: 1 = 500$

Each year thereafter the number of leaves was 40% more than the year before so that means

$Year \: 2 = 500(1+0.40) = 500\times 1.4\\$

For the third year the number of leaves increase 40% than the year before so that means

$Year \: 3 = 500\times 1.4(1+0.40) = 500 \times 1.4^{2}\\$

Similarly for fourth year,

$Year \: 4 = 500\times 1.4^{2}(1+0.40) = 500\times 1.4^{3}\\$

So we can clearly see the pattern here

Let f(n) be the number of leaves on the tree in Mohameds back yard in the nth year since he started tracking it then general recursive formula is

$f(n)= 500\times(1.4)^{n-1}\\$

This is the required recursive formula to find the number of leaves for the nth year.

Bonus:

Lets find out the number of leaves in the 10th year,

$f(10)= 500\times(1.4)^{10-1}\\\\f(10)= 500\times(1.4)^{9}\\\\f(10)= 500\times20.66\\\\f(10)= 10330$

So there will be 10330 leaves in the 10th year.

3 0

3 years ago

Read 2 more answers

Please help me with my homework.

Blababa [14]

Answer:

-3x + 6

Step-by-step explanation:

-3(x - 2) to find the equivalent of this expression, we need to multiply inside the parenthesis with -3 (with both x and -2)

-3(x - 2 = -3x + 6 (two negative expressions multiplied results in positive)

3 0

3 years ago

Sam collected six leaves to feed to his caterpillar collection if he wanted to split the leaves among seven cages how much shoul

Kazeer [188]

6÷7 is 0.857. maybe?

3 0

3 years ago