Para la fiesta de fin de cursos, debemos adornar 2 muros del salón y tenemos los siguientes materiales: 3m de listón blanco, 1.5 de listón dorado y 1/3 de flores. Para repartir el material en cantidades iguales en cada muro, vamos hacer lo siguiente: LISTÓN BLANCO, 3m entre 2, hacemos una división. 1m=2/2 entonces 3m=6/2, 6/2 entre 2=3/2m, tenemos 3/2m de listón blanco para cada muro. 1.5m entre 2, dividimos 1m en medios=2/2, 2/2 entre 2= 1/2m ahora: 1/2m en cuartos= 2/4, 2/4 entre 2= 1/4m falta sumar 2/4+1/4=3/4m. Tenemos 3/4m de listón dorado para cada muro. Resuelve el de las flores

The answer is below.

Step-by-step explanation:

For the end of courses party, we must adorn 2 walls of the living room and we have the following materials: 3m white ribbon, 1.5 gold slat and 1/3 flower. To distribute the material in equal quantities on each wall, let's do the following: WHITE LIST, 3m between 2, we do a division. 1m-2/2 then 3m-6/2, 6/2 between 2-3/2m, we have 3/2m of white slat for each wall. 1.5m between 2, We divide 1m into media 2/2, 2/2 between 2 x 1/2m now: 1/2m in quarters 2/4, 2/4 between 2o 1/4m need to add 2/4+1/4-3/4m. We have 3/4m of golden slat for each wall. Solve the flower one

it's

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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