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3 years ago

Solve the equation by completing the square: 3x^2 – 9x = -6

2 answers:

6 0

So my step one would be to add the six over and get 3x squared -9x+6=0, then divide by 3 to get x squared-3x+2=0. The next step is seeing what two numbers multiply to 2 but also add to negative three. The two numbers are -1 and -2. So then you are going to make two equations, one being x-1=0 and the other being x-2=0. Solve both to get X=1 and X=2, check your answers in the original equation. Hope this helps!

aleksandrvk [35]3 years ago

6 0

3x^2 - 9x = -6....divide everything by 3

x^2 - 3x = -2
x^2 - 3x + 2.25 = -2 + 2.25
(x - 1.5)^2 = 0.25
x - 1.5 = (+-) sqrt 0.25
x = 1.5 (+-) 0.5

x = 1.5 + 0.5 = 2 <===
x = 1.5 - 0.5 = 1 <===

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Convert 16/3 into a decimal form

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16/3 in decimal form = 3 ˙ 5.

Step-by-step explanation:

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3 years ago

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Find x. Round your answer to the nearest integer.

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3 years ago

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

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3 years ago

2m + 14 = -3?I need the steps and answer

Inessa [10]

2m+14=-3
First, subtract 14 from both sides to make the problem simplifer.
2m+14-14=-3-14
2m=-17
Divide both sides by 2 to solve the problem.
2m/2=-17/2
m=8.5
Hope it helps. :D

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3 years ago

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what are the possible numbers of positive, negative, and complex zeros of f(x)=x^6+x^5+x^4+4x^3-12x^2+12?

Lerok [7]

The number of zeros are; Positive; 2 or 0; Negative: 4 or 2 or 0; Complex; 0, 2, 4, 6

<h3>How to find the zeros of a Polynomial?</h3>

We are given the polynomial function as;

f(x) = x⁶ + x⁵ + x⁴ + 4x³ - 12x² + 12

Use Descartes Rule rule of signs allows us to determine the possible number of positive, negative, and complex roots.

For the function f(x), the number of sign changes that occur from term to term will give us the number of positive possible roots.

For the function f(-x), the number of sign changes that occur from term to term will give us the number of negative possible roots.

The numbers of complex roots is the difference between the degree of function and total number of positive and negative roots.

For f(x), the signs change 2 times. Thus, there can only be a possibility of 2 positive roots

For f(-x), we have;

x⁶ - x⁵ + x⁴ - 4x³ - 12x² + 12

Thus, the sign changes 4 times and so we have a possibility of 4 negative roots.

Read more about Descartes rule of polynomials at; brainly.com/question/12006853

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2 years ago