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Akimi4 [234]
4 years ago
10

Here is Triangle A. Lin created a scaled copy of Triangle A with an area of 72 square units. a. How many times larger is the are

a of the scaled copy compared to that of Triangle A? b. What scale factor did Lin apply to the Triangle A to create the copy?

Mathematics
1 answer:
PSYCHO15rus [73]4 years ago
7 0

Answer:

The answer is below

Step-by-step explanation:

a) Triangle A is attached in the image below.

The base of triangle A is 3 units and its height is 3 units. The area of a triangle  is given as:

Area = (1/2) × base × height

Area of triangle A = (1/2) × base × height = (1/2) × 3 × 3 = 4.5 unit²

Area of the scaled copy = 72 unit²

Ratio of area = Area of the scaled copy / Area of triangle A = 72 unit² / 4.5 unit² = 16

Hence the scaled copy area is 16 times larger than that of triangle A.

b) For the scaled copy:

Area of the scaled copy = (1/2) × base × height = 72 unit²

base × height = 144

Since the base and height are equal

base² = 144

base = 12, also height = 12

Base of scaled copy = 12 = 4 × base of triangle A

Therefore the scale factor used is 4

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3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
julia-pushkina [17]

With respect to the <em>parent</em> function y = cos x, we have these function according to the order of the graphs:

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  2. cos (1/2)x
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<h3>How to find the functions corresponding to each graph based on the period of a function</h3>

Mathematically speaking, cosines are <em>periodic</em> functions and period (T) is the distance in the <em>horizontal</em> axis such that f(t + T) = f(t). The period of the <em>parent</em> <em>cosine</em> function is 2π. Period in <em>daughter</em> functions may vary by means of the following form:

y = cos Ax     (1)

Where:

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Please notice that <em>daughter</em> functions report periods <em>longer</em> than in the <em>parent</em> function for 0 < A < 1 and a <em>shorter</em> one for A > 1. In consequence, these functions appear in the following order:

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6 0
2 years ago
Anne wants to fill 12 hanging baskets with compost. Each hanging basket is a hemisphere of diameter 40 cm.
murzikaleks [220]

Answer:

no

Step-by-step explanation:

Steps to answering this question

  1. determine the volume of the 12 basket

volume of a hemisphere = (2/3)πr^3

n = 22/7

r = radius

the diameter is the straight line that passes through the centre of a circle and touches the two edges of the circle.

A radius is half of the diameter

radius = 40/2 = 20 cm

volume of one hemisphere = (2/3) x (22/7) x (20^3) = 16,761.90 cm^3

volume of the 12 baskets = 16,761.90 cm^3 x 12 = 201,142.86 cm^

2. convert the litres of compost to cm and multiply by the total bags of compost

1 litre = 1000cm

1 bag of compost = 50 x 1000 = 50,000

4 bags of compost = 50,000 x 4 = 200,000 cm

3. compare which figure is higher. the figure gotten in step 1 or 2

201,142.86 cm^3 is greater than 200,000

there is no enough compost

473691.4

273691.4

3 0
3 years ago
See ss below<br>please show all work ​
____ [38]
Answer:

x = 117km

Step-by-step explanation:

a1 = 18 km

q = 90% = 90/100 = 0.9


a2 = a1 • q = 18 • 0.9 = 81/5 = 16.2 km

a3 = a2 • q = 16.2 • 0.9 = 729/50 = 14.58 km

n = 10

x = a1 • q^n - 1/q - 1 = 18 • 0.9^10 - 1/0.9 - 1 = 117km



7 0
3 years ago
Read 2 more answers
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