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3 years ago

Factorise X2 + 3√3x + 6. Please help me with this it’s urgent

Mathematics

1 answer:

ser-zykov [4K]3 years ago

6 0

Answer:

$(x+2\sqrt{3} )(x+\sqrt{3} )$

Step-by-step explanation:

$x^2 + 3\sqrt{3}x + 6\\=> x^2 + \sqrt{3}x + 2\sqrt{3}x +6\\=> x( x+\sqrt{3}) + 2\sqrt{3}(x+\sqrt{3} )\\ => (x+2\sqrt{3} )(x+\sqrt{3} )$

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4 years ago

Read 2 more answers

Choose all the expressions that are equal t o 3/4 × 9 . *

ozzi

Answer:

3÷4×9, 9×3/4

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3 years ago

Wesley estimated 5.82 - 4.21 to be about 2. Is this an overestimate or underestimate? Explain.

muminat

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3 years ago

Surface integrals using a parametric description. evaluate the surface integral \int \int_{s} f(x,y,z)dS using a parametric desc

DiKsa [7]

You can parameterize $S$ using spherical coordinates by

$\vec s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial\vec v}\times\dfrac{\partial\vec s}{\partial\vec u}=36\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle$

(I use $\vec s_v\times\vec s_u$ to avoid negative signs. The orientation of the normal vector doesn't matter for a scalar surface integral; you could just as easily use $\vec s_u\times\vec s_v=-(\vec s_v\times\vec s_u)$ .)

Then

$f(x,y,z)=f(6\cos u\sin v,6\sin u\sin v,6\cos v)=36\sin^2v$

and the integral of $f$ over $S$ is

$\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\int_0^{\pi/2}\int_0^{2\pi}36\sin^2v\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$