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4 years ago

15

Shen’s mother gave him $10 to go buy groceries. He picked up a gallon of milk for $3.49, three pounds of oranges that cost $1.14

per pound, and a box of cereal for $3.46. Shen used front-end estimation to determine whether he had enough money for the groceries. How much did he estimate the total cost would be?
He estimated that he had about $7 worth of groceries.
He estimated that he had about $9 worth of groceries.
He estimated that he had about $10.30 worth of groceries.
He estimated that he had about $10.40 worth of groceries.

1 answer:

garik1379 [7]4 years ago

7 0

Answer:

He estimated that he had about $10.40 worth of groceries.

Step-by-step explanation:

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Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

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$B(2, 6) = (x_2, y_2)$

$AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}$

$AB = \sqrt{(8)^2 + (4)^2}$

$AB = \sqrt{64 + 16} = \sqrt{80}$

$AB = 8.9$ (nearest tenth)

Distance between B(2, 6) and C(7, 1):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(2, 6) = (x_1, y_1)$

$C(7, 1) = (x_2, y_2)$

$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

$BC = \sqrt{(5)^2 + (-5)^2}$

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$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

$CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}$

$CD = \sqrt{(-4)^2 + (-6)^2}$

$CD = \sqrt{16 + 36} = \sqrt{52}$

$CD = 7.2$ (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

$DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

$DA = \sqrt{(-9)^2 + (7)^2}$

$DA = \sqrt{81 + 49} = \sqrt{130}$

$DA = 11.4$ (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

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