Need help with homework will give crown

1. You are standing on top of a ridge using a clinometer trying to figure out how far your dog has wandered off in

olga_2 [115]

Given:

It is given that the ridge is 360 inches tall.

Assumptions:

Assume you are 170.1 cm tall which equals 67 inches tall, the height from your eye to the floor is $360+67 = 427$ inches.

The distance from your eye level to the bottom of the ridge is 427 inches.

Assume the angle A is 60°.

To find the distance from you to your dog.

Solution:

A right-angled triangle can be formed where the angle is 60°, the distance between you and the dog is the hypotenuse of the triangle and your height from the floor is the adjacent side of the triangle.

Assume the hypotenuse of the triangle measures x inches.

To determine the length of the hypotenuse, we determine the cos of the angle.

$cos \theta = \frac{adjacent side}{hypotenuse} .$

$cos A = \frac{427}{x} , x = \frac{427}{cos60} , cos60=0.5.$

$x=\frac{427}{0.5} = 854.$

So if the ridge is 360 inches tall and you are 67 inches tall and the angle A is 60°, the distance between your dog and you is 854 inches.

3 0

2 years ago

3k-7/5 = 16 what is does k equal?

Gnom [1K]

3k=16+7/5
3k=87/5
k=29/5

3 0

2 years ago

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ x.

zepelin [54]

Answer:

$\mathbf{3x^2y^3+2xy^4=C}$

Step-by-step explanation:

From the differential equation given:

$6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0$

The equation above can be re-written as:

$6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0$

$(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0$

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

$\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}$

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

$M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3$

So;

$\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3$

$\dfrac{\partial N}{\partial y }$

Let's Integrate $\dfrac{\partial F}{\partial x}= M(x,y)$ with respect to x

Then;

$F(x,y) = \int (6xy^3 +2y^4) \ dx$

$F(x,y) = 3x^2 y^3 +2xy^4 +g(y)$

Now, we will have to differentiate the above equation with respect to y and set $\dfrac{\partial F}{\partial x}= N(x,y)$ ; we have:

$\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1$