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nataly862011 [7]

3 years ago

6

Find the altitude of a regular tetrahedron whose volume is 486 square root of 2 cm^3.

1 answer:

myrzilka [38]3 years ago

6 0

Using the formula:
V = a^3 / 6 square root of 2
h = square root 2/3 x a

h = 2/3 x 3%
V^1/3 = 2/3 x 3% x 486^2/3 = <span>13.09348
</span>
I hope this answer helps. Looking forward to help you again. Have a great day!

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Luden [163]

To find the equation of this line in slope-intercept form (y = mx + b, where m is its slope and b is its y-intercept), we naturally need the slope and the y-intercept. We can see that the line intersects the y-axis at the point (0, 4) so our y-intercept is 4, and the line rises 4 along the y-axis for every 2 it runs along the x-axis, so its slope is 4/2 = 2. With this in mind, we can write the line's equation as

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Answer:

Step-by-step explanation:

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Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

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