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3 years ago

60/8+X + 60/8-X = 16

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

3 0

Answer:

Step-by-step explanation:

$Domain:\\8+x\neq0\ \wedge\ 8-x\neq0\\\boxed{D:x\neq-8\ \wedge\ x\neq8}\\\\\dfrac{60}{8+x}+\dfrac{60}{8-x}=16\\\\\dfrac{60(8-x)}{(8+x)(8-x)}+\dfrac{60(8+x)}{(8+x)(8-x)}=16\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{60(8-x)+60(8+x)}{8^2-x^2}=16\qquad\text{use the distributive property}\\\\\dfrac{(60)(8)+(60)(-x)+(60)(8)+(60)(x)}{64-x^2}=16\\\\\dfrac{480-60x+480+60x}{64-x^2}=16\qquad\text{combine like terms}\\\\\dfrac{(480+480)+(-60x+60x)}{64-x^2}=16\\\\\dfrac{960}{64-x^2}=16$

$\dfrac{960}{64-x^2}=\dfrac{16}{1}\qquad\text{cross multiply}\\\\16(64-x^2)=(960)(1)\qquad\text{use the distributive property}\\\\(16)(64)+(16)(-x^2)=960\\\\1024-16x^2=960\qquad\text{subtract 1024 from both sides}\\\\1024-1024-16x^2=960-1024\\\\-16x^2=-64\qquad\text{divide both sides by (-16)}\\\\\dfrac{-16x^2}{-16}=\dfrac{-64}{-16}\\\\x^2=4\to x=\pm\sqrt4\\\\x=\pm2\in D$

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Answer:

The area of a parallelogram is 360 in.²

Step-by-step explanation:

Where DG = GH

GP = 12 in.

AB = 39 in.

∠DAB + ∠ABC = 180° (Adjacent angles of a parallelogram)

Whereby ∠DAB is bisected by AG and ∠ABC is bisected by BH

Therefore, ∠GAB + ∠HBA = 90°

Hence, ∠BPA = 90° (Sum of interior angles of a triangle)

$cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{AP}{39} = \dfrac{GP}{GH} =\dfrac{12}{GH}$

We note that ∠AGD = ∠GAB (Alternate angles of parallel lines)

∴ ∠AGD = ∠AGD since ∠AGD = ∠GAB (Bisected angle)

Hence AD = DG (Side length of isosceles triangle)

The bisector of ∠ADG is parallel to BH and will bisect AG at point Q

Hence ΔDAQ ≅ ΔDGQ ≅ ΔGPH and AQ = QG = GP

Hence, AP = 3 × GP = 3 × 12 = 36

$cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{36}{39}$

$\angle GAB = cos^{-1} \left (\dfrac{36}{39} \right )$

∠GAB = 22.62°

$cos(\angle GAB) = \dfrac{36}{39} = \dfrac{12}{GH}$

$GH = \dfrac{39}{36} \times {12}$

GH = 13 in.

∴ AD 13 in.

BP = 39 × sin(22.62°) = 15 in.

GH = √(GP² + HP²)

∠DAB = 2 × 22.62° = 45.24°

The height of the parallelogram = AD × sin(∠DAB) = 13 × sin(45.24°)

The height of the parallelogram = 120/13 = 9.23 in.

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