134d) Graph the hyperbola, identify the center, vertices, foci, and equations of the asymptotes 49y^2−25x^2−196y−200x−1429=0

IF YOU ARE GOOD AT MATH, PLEASE HEAD OVER TO MY PROFILE, IN THE QUESTIONS TAB AND ANSWER THE QUESTIONS STARTING FROM THE ONE MARKED 132B!

best answers get marked 'brainliest'!

Thank you!

Step-by-step explanation:

49y^2 - 25x^2 - 196y - 200x = 1429

49y^2-196y+196 - 25x^2-200x-400 = 1429+196-400

49(y-2)^2 - 25(x+4)^2 = 1225

(y-2)^2/25 - (x-4)^2/49 = 1

So now you know that you have an hyperbola with

center at (-4,2)

foci at (-4,2±)

vertices at (-4,2±5)

asymptotes 7y±20 = ∓5x+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

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet