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3 years ago

Find the area of this parallelogram in (cm2).

Mathematics

1 answer:

tester [92]3 years ago

6 0

The answer to your question is 260(cm2)

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Alice stand at point A and looks at the top of a 17.8 m tree TB, such that her line of sight makes an angle 38° with the horizon

Nesterboy [21]

Answer:

Step-by-step explanation:

let the horizontal distance be x

(17.8-1.5)/x = tan 38

or x = 16.3/tan38

or x = 20.863

4 0

2 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

10. A young child is 44 months old. Find the age of the baby in years as a mixed

viktelen [127]

Step-by-step explanation:

44 month old

12 months=1 year

44÷12=3 2/3

3 2/3 years

5 0

3 years ago

Read 2 more answers

Which could be the entire interval over which the function,

Marrrta [24]

Answer:

(-2, 1)

Step-by-step explanation:

Hi Juliadayx! How are you?

Well, maybe you already heard about the domain and the range of a function.

The domain is the set of values that the independent variable can take (usually referred to as the letter "x"), while the range is the set of values that the dependent variable takes, which is called f(x) or function of x.

In this case, the exercise asks you to evaluate for which values of “x” (right column of the table), the function “f(x)” takes positive values (left column of the table), the positive values also include zero. And in this case you can see that the function f(x) is only positive for the values of "x": -2, -1, 0 and 1. Therefore, the answer is the entire interval (-2, 1).

I hope I've been helpful!

Regards!

3 0

3 years ago

How do I find the area?

nikitadnepr [17]

Answer:

The simplest (and most commonly used) area calculations are for squares and rectangles. To find the area of a rectangle, multiply its height by its width. For a square, you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.

5 0

3 years ago

Read 2 more answers