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3 years ago

What is the thousandths value in 62.407

Mathematics

1 answer:

den301095 [7]3 years ago

6 0

The thousandths value is 7.

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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 :

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

7 0

4 years ago

Plot the x- and y-intercepts to graph the equation.

nikdorinn [45]

Answer:

(13,12)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find cos θ given that cos 2θ = 5/6 and 0 ≤ θ < π/2. Give an exact answer

trasher [3.6K]

$\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
1-2sin^2(\theta)\\
\boxed{2cos^2(\theta)-1}
\end{cases}
\\ \quad \\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-------------------------------\\\\
$

$\bf cos(2\theta)=\cfrac{5}{6}\implies 2cos^2(\theta)-1=\cfrac{5}{6}\implies 2cos^2(\theta)=\cfrac{5}{6}+1
\\\\\\
2cos^2(\theta)=\cfrac{11}{6}\implies cos^2(\theta)=\cfrac{11}{12}\implies cos(\theta)=\pm\sqrt{\cfrac{11}{12}}$

now, bear in mind, the square root gives us +/- versions, so, which is it? well, we know the angle is in the range of "<span>0 ≤ θ < π/2", that simply means the 1st quadrant, so, we'll use the positive one then

</span> $\bf cos(\theta)=\cfrac{\sqrt{11}}{\sqrt{12}}\implies cos(\theta)=\cfrac{\sqrt{11}}{2\sqrt{3}}
\\\\\\
\textit{now, let's rationalize the denominator}
\\\\\\
\cfrac{\sqrt{11}}{2\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{11}\cdot \sqrt{3}}{2\sqrt{3^2}}\implies \cfrac{\sqrt{11\cdot 33}}{2\cdot 3}\implies \boxed{\cfrac{\sqrt{33}}{6}}$ <span>
</span>

3 0

4 years ago

Dustin is riding his bike from Logan, Utah to Jackson Hole, Wyoming in a bike race. He must ride at least 200

sashaice [31]

Answer:the answers is 88 1/2

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

An object travels along the x-axis so that its position after t seconds is given by x(t) = 2t2 – 5t – 18 for all times t such th

Degger [83]

the function is given, and it's value is where the object is ("how far to the right").

so as long as it rises (going more right), this will be apply.

in the screenshot I graphed the function. of course t is graphed as x and "along the x-axis" is graphed as y, but the pattern is the same anyways.

for the first 1.25 seconds the object goes to the left, and after that always to the right.

since we look at t to calculate x, t effectively takes the role of the important variable that is normally given to x. the calculation pattern are just the same. so let's find the lowest point of this function by calculating it out.

x(t) = 2t² – 5t – 18

x'(t) = 4t -5

x'(t) = 0

0 = 4t -5

5 = 4t

1.25 = t

plugging it into the second derivative

x''(t) = 4

x''(1.25) = 4

it's positive, so at t=1.25 there is a low point

(of course the second derivative is constant anyways.)

the object is traveling toward the right

the object is traveling toward the rightfor t > 1.25

7 0

3 years ago