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2 years ago

I need help with this, pls put step by step

Mathematics

1 answer:

choli [55]2 years ago

5 0

9/24 = 3/u

Cross multiply
24u = 27
U = 27/24

Simplify
U = 9/8 cm

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Divide (-2x-3+8x^2) by (1 + 4x)

Strike441 [17]

Answer:

8x^2-2x-3/1+4x

Step-by-step explanation:

It doesn't say any of your answers but i just used a calculator and that's what i got

Sorry if its wrong

8 0

3 years ago

What is the missing addend?<br> 52,835 +<br> = 87,216

posledela

Answer: 34,381

Step-by-step explanation:

You have to subtract 52,835 by the sum 87,216 which will get you the answer of 34,381

6 0

3 years ago

Read 2 more answers

What is the arc length of the arc sub tended in a circle with radius 6 and an angle of 7 pi/ 8

KengaRu [80]

Answer:

Formula for the Arc length is given by:

$\text{Arc length} = 2 \pi r \cdot \frac{\theta}{360^{\circ}}$

As per the statement:

radius of circle(r) = 6 units

Angle ( $\theta$ ) = $\frac{7 \pi}{8}$ radian

Use conversion:

$1 rad = \frac{180}{\pi}$

$\frac{7 \pi}{8}$ = $\frac{180}{\pi} \cdot \frac{7 \pi}{8} = \frac{1260}{8} = 157.5^{\circ}$

then;

substitute these given values we have;

Use value of $\pi = 3.14$

$\text{Arc length} = 2\cdot 3.14 \cdot (6) \cdot \frac{157.5^{\circ}}{360^{\circ}}$

$\text{Arc length} = 2\cdot 3.14 \cdot (6) \cdot 0.4375$

Simplify:

$\text{Arc length}=16.485$

Therefore, the arc length of the arc substended in a circle with radius 6 units an angle of 7 pi/8 is 16.485 units

7 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

Which binomial is a difference of squares?

GenaCL600 [577]

D) 36x^2-81
both 36x^2 and -81 are perfect squares
difference means subtraction

7 0

3 years ago

I need help with this, pls put step by step ​

I need help with this, pls put step by step