All categories

3 years ago

50 (x+ 10) = 240 Whats the value for x

Mathematics

2 answers:

Katen [24]3 years ago

3 0

Answer:

-26/5 or -5.2

Step-by-step explanation:

First, you need to expand the parentheses:

50x+500=240

Now, subtract 500 from both sides:

50x=-260

Now, divide both sides by 50

x=-26/5=-5.2

Hope this helps!

weeeeeb [17]3 years ago

3 0

ANSWER:

x=-14.8

SOLVING STEPS:

1. Divide both sides by 50
2. Subtract 10 from both sides

x+10=4.8
x=-14.8

You might be interested in

Β<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20" id="TexFormula1" title=" \int\limits^a_b

4vir4ik [10]

Hint:- integral of x is x^2 / 2

5 0

3 years ago

Eddie is reading a novel for English class. He read 173 out of 480 pages. About what percent of the book has he read? Round to t

tester [92]

Answer:

the answer would be 36.04,

Step-by-step explanation:

im gonna just leave this blank

4 0

3 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

HELP PLEASE ILL GIVE BRAINLIEST!

yulyashka [42]

It should be 3 because you are going to be using the formation of 3√ 3•3•3

4 0

2 years ago

?????????????? So like anyone know this?

Ivanshal [37]

Not sure sorry I’m not good at math need help with something else..

8 0

3 years ago