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3 years ago

Find the slip of the life that passes through the two points (7,-10) and (-6,16)

Mathematics

1 answer:

Anastaziya [24]3 years ago

4 0

Answer: Slope = -2

Hope this helps!!! Good luck!!! ;)

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Solve the initial value problems.

slavikrds [6]

Both equations are linear, so I'll use the integrating factor method.

The first ODE

$xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0$

has integrating factor

$\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x$

In the original equation, multiply both sides by eˣ :

$xe^x y' + (x+1) e^x y = 0$

Observe that

d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ

so that the left side is the derivative of a product, namely

$\left(xe^xy\right)' = 0$

Integrate both sides with respect to x :

$\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx$

$xe^xy = C$

Solve for y :

$y = \dfrac{C}{xe^x}$

Use the given initial condition to solve for C. When x = 1, y = 2, so

$2 = \dfrac{C}{1\cdot e^1} \implies C = 2e$

Then the particular solution is

$\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}$

The second ODE

$(1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0$

has integrating factor

$\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}$

Multiply both sides of the equation by 1/(1 + x²) :

$\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0$

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

$\left(\dfrac1{1+x^2}y\right)' = 0$

$\dfrac1{1+x^2}y = C$

$y = C(1 + x^2)$

When x = 0, y = 3, so

$3 = C(1+0^2) \implies C=3$

$\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}$

7 0

2 years ago

Mike hiked to big bear lake in 4.5 hours at an average rate of 3 1/5 miles per hour. Pedro hiked the same distance at a rate of

Goryan [66]

It took Pedro 4 hours.

3 0

3 years ago

What is the slope for (-19, -6), (15, 16)

Licemer1 [7]

Answer:

11/17≈0.647059

Step-by-step explanation:

4 0

3 years ago

After several shares of the company's stock were sold, a profit of $1.320 was earned. The profit was 15%

Rom4ik [11]

Answer:

$8.8

Step-by-step explanation:

Let the original price of shares be $x$

Profit earned = $1.320

percentage profit = 15%

∴ $\frac{15x}{100} = 1.320\\15x = 132.0\\x = \frac{132}{15} \\x = 8.8$

Original price of shares were = $8.8

4 0

3 years ago

Please help I will give brainliest! I just need 22 and 23 please! Thank you! (And it’s related to ordered pairs by the way!)

cluponka [151]

Answer:

22. B1, G1 (from left to right)

23. E5

Step-by-step explanation:

7 0

3 years ago