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3 years ago

Please Help !!!!

Mathematics

2 answers:

ikadub [295]3 years ago

5 0

It is A. 2 because it is the same as the highest degree

Nat2105 [25]3 years ago

4 0

Answer:

Polynomial has three roots.

B is the correct option.

Step-by-step explanation:

We have been given the function $y=(x-2)(x+3)^2$

We know that the number of roots or zeros of a polynomial is equal to the degree of the polynomial.

Let us simplify the given function and find the degree.

$y=(x-2)(x+3)^2\\\\y=(x-2)(x^2+6x+9)\\\\y=x^3+6x^2+9x-2x^2-12x-18\\\\y=x^3+4x^2-3x-18$

Since, the degree of the polynomial is 3.

Hence, it must have three zeros or roots.

B is the correct option.

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How is 0.136¯¯¯¯ written as a fraction in simplest form?

drek231 [11]

Answer:

$\frac{3}{22}$

Step-by-step explanation:

the bar above 36 means that the digits 36 are being repeated

we require 2 equations with the repeating digits placed after the decimal point.

let x = 0.13636.. ( multiply both sides by 10 and 1000 )

10x = 1.3636... (1)

1000x = 136.3636... (2)

subtract (1) from (2) thus eliminating the repeating digits

990x = 135 ( divide both sides by 990 )

x = $\frac{135}{990}$ = $\frac{3}{22}$ ← in simplest form

5 0

2 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$