All categories

3 years ago

Could u plz answer this2x-7=133x+4=25

Mathematics

1 answer:

sashaice [31]3 years ago

5 0

Answer:

x=10

Step-by-step explanation:

You might be interested in

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

7 0

3 years ago

I don't understand this nonsense !

mel-nik [20]

Well first you do 6÷3=2. Then you do 2*5= 10. Also do 20+10=30-15=15. So your answer is 15

5 0

3 years ago

Hey! I am stuck on this question anyone mind helping me? and tysm for the people who helped me! <3 have a nice day!

Andrej [43]

Answer: 166 2/3m

Step-by-step explanation:

Area of a rectangle = length times width

Length: 16 2/3

To find the width, multiply it by 3/5.

w = 16 2/3 · 3/5

w = 10

Then, multiply the length by the width.

a = l · w

a = 16 2/3 · 10

a = 166 2/3m

4 0

3 years ago

3) Graph by making a<br> table.<br> y=x

Zanzabum

Answer:

See attachment for graph

Step-by-step explanation:

Given

$y = x$

Required

Plot the graph

First, we need to make a table.

To do this, we assume values for x

When $x = 1$

$y = x = 1$

$(x_1,y_1) = (1,1)$

When $x = 2$

$y = x = 2$

$(x_2,y_2) = (2,2)$

When $x = 3$

$y = x = 3$

$(x_3,y_3) = (3,3)$

When $x = 4$

$y = x = 4$

$(x_4,y_4) = (4,4)$

So, the table is:

$x || 1 || 2 || 3 || 4$

$y || 1 || 2 || 3 || 4$

<em>See attachment for graph</em>

8 0

3 years ago

Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal place

gayaneshka [121]

Answer:

The Riemann Sum for $\int\limits^5_4 {x^2+4} \, dx$ with n = 4 using midpoints is about 24.328125.

Step-by-step explanation:

We want to find the Riemann Sum for $\int\limits^5_4 {x^2+4} \, dx$ with n = 4 using midpoints.

The Midpoint Sum uses the midpoints of a sub-interval:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We know that a = 4, b = 5, n = 4.

Therefore, $\Delta{x}=\frac{5-4}{4}=\frac{1}{4}$

Divide the interval [4, 5] into n = 4 sub-intervals of length $\Delta{x}=\frac{1}{4}$

$\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right]$

Now, we just evaluate the function at the midpoints:

$f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625$

$f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625$

$f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625$

$f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625$

Finally, use the Midpoint Sum formula

$\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125$

This is the sketch of the function and the approximating rectangles.

5 0

4 years ago

Could u plz answer this2x-7=133x+4=25​

Could u plz answer this2x-7=133x+4=25