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3 years ago

11

Round 35,894 to the greatest place

2 answers:

laiz [17]3 years ago

7 0

35,894 rounded to the greatest place is 36,000

amid [387]3 years ago

4 0

Hoping that I rounded it to the right place.
It would be 35,900

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Jet001 [13]

Answer:

15 inches

Step-by-step explanation:

<u>The ratio of sides in 30º-60º-90º triangle is:</u>

a : b : c = 1 : √3 : 2

<u>If a is 5√3, the length of b is:</u>

b = a√3 = 5√3 *√3 = 5*3 = 15

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2 years ago

describe how the location of the vertex of the angle determines the process of finding the angle measures associated with that a

LekaFEV [45]

Answer:

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Step-by-step explanation:

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3 years ago

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Semmy [17]

Answer:

Step-by-step explanation:

54, $1\frac{2}{3}$ , 1, 0.5, 0.03, 0, - 1, - $1\frac{3}{4}$ , - 4, - 103

5 0

3 years ago

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 :

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

7 0

3 years ago

Expand the following:<br> a) x(x + 2)<br> b) x(2x - 5)<br> c) 2x(3x + 4)<br> d) 6x(x - 2)

TEA [102]

Answer:

<h2> $a. \: {x}^{2} + 2$ </h2>

<h2> </h2><h2> $b. \: 2 {x}^{2} - 5x$ </h2><h2> </h2><h2> $c. \: 6 {x}^{2} + 8x$ </h2><h3> </h3><h3> $d. \: 6 {x}^{2} - 12x$ </h3>

solution,

$a. \: x(x + 2) \\ \: \: = x \times x + 2 \times x \\ \: \: = {x}^{2} + 2x$

$b . \: x(2x - 5) \\ \: = x \times 2x - x \times 5 \\ \: \: = 2 {x}^{2} - 5x$

$c. \: 2x(3x + 4) \\ \: = 2x \times 3x + 2x \times 4 \\ \: \: = 6 {x}^{2} + 8x$

$d. \: 6x(x - 2) \\ \: \: = 6x \times x - 6x \times 2 \\ \: \: = 6 {x}^{2} - 12x$

<h2> </h2><h2 />

Hope this helps...

Good luck on your assignment..

4 0

3 years ago