Just help me out with this polynomial i need help { -m^{2} + 6} + { -4m^{2} + 7m + 2} =

Mathematics

1 answer:

PolarNik [594]3 years ago

7 0

Answer:

$-5m^2+7m+8$

Step-by-step explanation:

Here, we add up two polynomials shown.

The polynomials are:

$[-m^2 + 6]+[-4m^2 +7m + 2]$

In order to add up the 2 polynomials shown, we have to see the "like terms" and add them up.

We add up the " $m^2$ " terms and the constant (number) terms. There is one term with "m", so we leave it like that. Let's add up. Shown below:\

$[-m^2 + 6]+[-4m^2 +7m + 2]\\=-m^2-4m^2+6+2+7m\\=-5m^2+7m+8$

This is the sum of the 2 polynomials shown: $-5m^2+7m+8$

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1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$