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Alina [70]
2 years ago
11

How do you write forty and twenty-eight hundredths in standard form

Mathematics
1 answer:
alisha [4.7K]2 years ago
5 0
The answer will be 40.28
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Last week at the lake it was 23 °C. Yesterday it was 13 degrees warmer. What was the temperature yesterday?
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23+13
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What is an algebraic expression for the word phrase the product of 11 and a number s?
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For the parallelogram ABCD the extensions of the angle bisectors AG and BH intersect at point P. Find the area of the parallelog
natita [175]

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.

The area of a parallelogram = Base × Height = (120/13) × 39 = 360 in.²

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Analyze the diagram below and complete the instructions that follow.
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