drag each label to the correct location on the equation. each label can be used more than once, but not all labels will be used.
find the standard form of the equation of the hyperbola x2 − 4y2 − 2x + 16y − 31 = 0. place the signs and values in the correct places in the standard form of the equation.

(x - 1)²/4² - (y - 2)²/2² = 1 ⇒ The bold labels are the choices

Step-by-step explanation:

* Lets explain how to solve this problem

- The equation of the hyperbola is x² - 4y² - 2x + 16y - 31 = 0

- The standard form of the equation of hyperbola is

  (x - h)²/a² - (y - k)²/b² = 1 where a > b

- So lets collect x in a bracket and make it a completing square and

  also collect y in a bracket and make it a completing square

∵ x² - 4y² - 2x + 16y - 31 = 0

∴ (x² - 2x) + (-4y² + 16y) - 31 = 0

- Take from the second bracket -4 as a common factor

∴ (x² - 2x) + -4(y² - 4y) - 31 = 0

∴ (x² - 2x) - 4(y² - 4y) - 31 = 0

- Lets make (x² - 2x) completing square

∵ √x² = x

∴ The 1st term in the bracket is x

∵ 2x ÷ 2 = x

∴ The product of the 1st term and the 2nd term is x

∵ The 1st term is x

∴ the second term = x ÷ x = 1

∴ The bracket is (x - 1)²

∵  (x - 1)² = (x² - 2x + 1)

∴ To complete the square add 1 to the bracket and subtract 1 out

   the bracket to keep the equation as it

∴ (x² - 2x + 1) - 1

- We will do the same withe bracket of y

- Lets make 4(y² - 4y) completing square

∵ √y² = y

∴ The 1st term in the bracket is x

∵ 4y ÷ 2 = 2y

∴ The product of the 1st term and the 2nd term is 2y

∵ The 1st term is y

∴ the second term = 2y ÷ y = 2

∴ The bracket is 4(y - 2)²

∵ 4(y - 2)² = 4(y² - 4y + 4)

∴ To complete the square add 4 to the bracket and subtract 4 out

   the bracket to keep the equation as it

∴ 4[y² - 4y + 4) - 4]

- Lets put the equation after making the completing square

∴ (x - 1)² - 1 - 4[(y - 2)² - 4] - 31 = 0 ⇒ simplify

∴ (x - 1)² - 1 - 4(y - 2)² + 16 - 31 = 0 ⇒ add the numerical terms

∴ (x - 1)² - 4(y - 2)² - 16 = 0 ⇒ add 14 to both sides

∴ (x - 1)² - 4(y - 2)² = 16 ⇒ divide both sides by 16

∴ (x - 1)²/16 - (y - 2)²/4 = 1

∵ 16 = (4)² and 4 = (2)²

∴ The standard form of the equation of the hyperbola is

   (x - 1)²/4² - (y - 2)²/2² = 1

b 1: 45 am

step-by-step explanation:

its that because if it was in the pm it would be 1345

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)