Mathematics, 03.10.2019 02:00 Imthiccasf

Consider the surface given by
2x^2+5y^2+z^2=4.
the surface is sliced by planes of the form x=a, y=b or z=c for values of a, b, c given below. how do i match the resulting slices (in other words, the corresponding traces) with their correct plots?

-- also the slices are:
1. z= 1.5
2. x= 1
3. z=0
4. y=0

Answers: 1

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Answers

This is just simple. For example you have a plane of the form x=a, then you just substitute x with a, and you'll get an equation with y and z only, hence you have a 2-d trace of the intersection. It is just similar for y=b and z=c.

(1) At z=1.5, 2x^2 + 5y^2 + 1.5^2 = 4
2x^2 + 5y^2 = 1.75
Now you have an ellipse in the z=1.5 plane as your trace.

(2) At x=1, 2(1)^2 + 5y^2 + z^2 = 4
5y^2 + z^2 = 2
Now you have an ellipse in the x=1 plane as your trace.

(3) At z=0, 2x^2 + 5y^2 + (0)^2 = 4
2x^2 + 5y^2 = 4
Now you have an ellipse in the z=0 plane as your trace.

(4) At y=0, 2x^2 + 5(0)^2 + z^2 = 4
2x^2 + z^2 = 4
Now you have an ellipse in the y=0 plane as your trace.

Answer from: Quest

$w\geq 14$

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

$2l+2w\geq 40$

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

$2\times 6+2w\geq 40$

now, we can solve for w

$12+2w\geq 40$

subtract both sides by 12

$12+2w-12\geq 40-12$

$2w\geq 28$

divide both sides by 2

and we get

$w\geq 14$

so,

the smallest possible measurement of the width is 14 feet

Answer from: Quest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

ver más en -

step-by-step explanation:

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Consider the surface given by
2x^2+5y^2+z^2=4.
the surface is sliced by planes of the fo...

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