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Mazyrski [523]
4 years ago
5

In February Lana got a puppy that weighed 13 kilograms. By October the puppy weighed 17.16 kilograms. What was the percent the p

uppy weight increased?
Mathematics
1 answer:
valentinak56 [21]4 years ago
3 0
(17.16-13)/13=0.32=32%
the percent of change=the amount of change/the original number
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A pair of sunglasses is on sale for $17. Last week, the sunglasses were marked at a regular price of $20. What was the percent o
padilas [110]

sunglasses are on sale this week for $17 and they were formally $20.

$20 x .15= $3

A= 15%

8 0
3 years ago
I need the answer please help !
Dmitry_Shevchenko [17]

Answer:

Step-by-step explanation:

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5 0
4 years ago
These liness pass through the same point
anyanavicka [17]

Answer:

All of them except B

Step-by-step explanation:

Bisecting is splitting a line in half

Parallel lines never touch each other

Intersecting lines intersect

Corresponding lines are 2 parallel lines with a line that is going through them

8 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
For a trapezoid, the following formula relates the area a, the two bases a and b, and the height h. a= g (a
Dovator [93]
A - area of a trapezoid;
a, b - two bases.
A = h * ( a + b ) / 2            / * 2  ( we have to multiply both sides by 2 )
2 A = h *( a + b )
2 A = h a + h b
h b = 2 A - h a
b = ( 2 A - h a ) / h
b = 2 A/h - a
Answer: D )
5 0
4 years ago
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