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Vinil7 [7]
3 years ago
14

In square ABCD, point M is the midpoint of side AB and point N is the midpoint of side BC. What is the ratio of the area of tria

ngle AMN to the area of square ABCD? Express your answer as a common fraction

Mathematics
2 answers:
ruslelena [56]3 years ago
7 0

Answer:

The ratio of the area of triangle AMN to the area of square ABCD is 1:8.

Step-by-step explanation:

Check attachment for solution

Phoenix [80]3 years ago
6 0

Answer:

Ratio of triangle AMN to square ABCD = 1/8

Step-by-step explanation:

area AMN = Area ABN - Area MNB

Where AB = BC = x

Area AMN = [ ½ *( x * ½x )] - [ ½( ½x * ½x)] = x²/8

Area of ABCD = x* x = x²

So therefore:

The ratio of area of triangle AMN to area of square ABCD would be

= (x²/8) / x²

=1/8

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5n – 7p + 3n = 25p

Combine like terms

8n - 7p = 25p

Add (7p) to both sides

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Simplifying

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8n / 8 = 32p / 8

Simplifying

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Which of the following is a non-prime factor of 70? Select one: A. 5 B. 7 C. 10 D. 15
Umnica [9.8K]

Answer:

C. 10

Step-by-step explanation:

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Plant A: A graph has time (weeks) on the x-axis, and height (inches) on the y-axis. A line goes through points (0, 3), (1, 4.8),
butalik [34]

Answer:

The correct option is;

No, The greater rate of change of Plant A will result in it being 0.9 inches taller in 6 weeks

Step-by-step explanation:

The parameters given are;

Plant A:

Weeks,    Height

0,               3

1,                4.8

2,               6.6

3,               8.4

The rate of change of height, H per week, t,\left (\dfrac{dH}{dt} \right )  for plant A per week is therefore;

\dfrac{dH}{dt} =  \dfrac{H_n - H_{(n-1)} }{t_n - t_{(n-1)}} = \dfrac{8.4 - 3 }{3 - 0} = \dfrac{5.4}{3} = 1.8 \ inches/week

Therefore we have;

H = 1.8 × t + 3

At week 6,

H = 3 + 6×1.8 = 13.8 inches

Plant B

Weeks,    Height

2,               7.3

3,               8.7

4,               10.1

The rate of change of height, H per week, t,\left (\dfrac{dH}{dt} \right )  for plant B per week is given as follows;

\dfrac{dH}{dt} =  \dfrac{H_n - H_{(n-1)} }{t_n - t_{(n-1)}} = \dfrac{10.1 - 7.3 }{4 - 2} = \dfrac{2.8}{2} = 1.4 \ inches/week

Therefore we have;

When t = 2, H = 7.3 hence, 7.3 = 2 × 1.4 + H₀

Where:

H₀ = H at t = 0

H₀ = 7.3 - 2 × 1.4 = 4.5

At week 6 we have;

H = 4.5 + 6×1.4 = 12.9 inches

Which indicates that Plant A will be 0.9 inches taller than Plant B at week 6.

The correct option is therefore;

No, The greater rate of change of Plant A will result in it being 0.9 inches taller in 6 weeks.

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