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Alinara [238K]
3 years ago
11

A bulletin board has an area of 1/ 12 yd and a length of 1/3 yard the bulletin board is divided into six equal sections right vi

sion expressions to solve each part and show your work
Mathematics
2 answers:
gulaghasi [49]3 years ago
8 0

Answer: The 6 sections has an area of 1/72 square yards

Step-by-step explanation:

Length= area / length

Length = 1/12 ÷ 1/3 = 1/12 × 3/1 = 1/4 yards

Since the width and length are different and the 6 squares must be arranged in a 2 × 3 pattern.

The long side is 1/3 yard

1/4 ÷ 2/1 = 1/4 × 1/2 = 1/8 yard

And the width is 1/3 ÷ 3/1 = 1/3 × 1/3 = 1/9 yard

The 6 sections are 1/8 × 1/9 and each one has an area of 1/8×1/9 = 1/72 square yard

nlexa [21]3 years ago
6 0

Answer:

Step-by-step explanation:

Given:

Area = 1/12 yd^2

Length = 1/3 yd

Area of a rectangle = length × width

Width = 1/12 ÷ 1/3

= 1/4

Since it is divided into 6 parts, it can be written as 2 × 3 pattern.

Since the length is the longer part, it eould be divided into 3 then the width into 2

Length of the section = 1/3 ÷ 3

= 1/9 yd

Width of the section = 1/4 ÷ 2

= 1/8 yd

Area of the section = length of the section × width of the section

= 1/8 × 1/9

= 1/72 yd^2

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Iteru [2.4K]

Answer:

-5/4

Step-by-step explanation:

4x-5y=10

-5y=-4x-10

-5y/-5=-4/-5-10/-5

y=4/5+2

perpendicular slope is -5/4

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Why is 6 1/2 equal to the square root of 6
Aleksandr-060686 [28]
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(3x + 1) &gt; -7+ 3x<br> Plz help me
Luda [366]

Answer:

0, -1

Step-by-step explanation:

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2 years ago
A trader made a profit of 24% by selling an article for 3,720cedis how much should he have sold it to make a profit of 48%​
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6 0
3 years ago
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.
konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

According to this theorem,

BC^2=BD\cdot AB.

Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then

x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}0.

Take positive value x. You get

AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.

2. According to the previous theorem,

AC^2=AD\cdot AB.

Then

AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

Answer: AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.

This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then

CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D

This means that you cannot find solutions of this equation. Then CD≠2 cm.

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3 years ago
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