1) Find the value of two numbers if their sum is 12 and their difference is 4.

Mathematics

1 answer:

Ivahew [28]2 years ago

4 0

Numbers are 4 and 8

Step-by-step explanation:

Step 1:

Let the numbers be x and y. Given that their sum is 12 and difference is 4. Form equations for this data.

⇒ x + y = 12 ------ (1)

⇒ x - y = 4 -------- (2)

Subtract eq (2) from (1)

⇒ 2x = 8

⇒ x = 4

Step 2:

Find y.

⇒ y = 12 - x = 12 - 4 = 8

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The family of solutions to the differential equation y ′ = −4xy3 is y = 1 √ C + 4x2 . Find the solution that satisfies the initi

slega [8]

Answer:

The correct option is 4

Step-by-step explanation:

The solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}$

Now for the initial condition the value of C is calculated as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(-2)=\frac{1}{\sqrt{C+4(-2)^2}}\\4=\frac{1}{\sqrt{C+4(4)}}\\4=\frac{1}{\sqrt{C+16}}\\16=\frac{1}{C+16}\\C+16=\frac{1}{16}\\C=\frac{1}{16}-16$

So the solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}$

Simplifying the equation as

$y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1-256+64x^2}{16}}}\\y(x)=\frac{\sqrt{16}}{\sqrt{{1-256+64x^2}}}\\y(x)=\frac{4}{\sqrt{{1+64(x^2-4)}}}$