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Ivanshal [37]
3 years ago
14

How do u make this 4/4?

Mathematics
1 answer:
Oksana_A [137]3 years ago
8 0

well, if look at the timetable, She got the 08:30am train from Aberystwyth and arrives on time at Shrewsbury, from the timetable we know she arrived at 10:17am, now she did some rigamarole and got back to the Train station at Shrewsbury 4 hours later.  Well, we know she arrived at 10:17am, if we add 4 hours to that that'll make it 1417 or namely 2:17pm.

well, the Train arrives at Shrewsbury a 14 19, or 2:19pm, she is there at 2:17pm, so she's really 2 minutes before the Train arrives at Shrewsbury, she's right on time, possibly with some munchies too.

14 19 > 14 17.

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Carlotta has 12 2/3 yd of ribbon to be divided among 19 girls in the cheerleading squad.how much ribbon will each girl get if it
trasher [3.6K]

Answer:

Each girl gets 2/3 yd of ribbon.

Step-by-step explanation:

12 2/3=38/3

(38/3)/19

(38/3)(1/19)

(2/3)(1/1)=2/3

4 0
3 years ago
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

7 0
3 years ago
Read 2 more answers
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