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finlep [7]
3 years ago
6

I don’t understand the problem or question

Mathematics
1 answer:
skad [1K]3 years ago
4 0

Answer:

Part A

a. Could be a parallelogram but cannot be a rectangle

Part B

c. 100°

Step-by-step explanation:

Part A

The polygon constructed by Andrew = Four sided polygon having no right angles

From the given four sided polygon (quadrilateral) options, the options which can have no right angles are the parallelogram, and rhombus

The rectangle, and square both have 4 right angles

Therefore, the polygon can be either a parallelogram or a rhombus, but cannot be either a rectangle or square

The correct option is therefore, 'His polygon could be a parallelogram but cannot be a rectangle'

Part B

The given parameters of the rhombus EFGH are;

Angle E = 3·x + 5, angle H = 4·x

The characteristics of a rhombus are;

The sum of the interior angles of a rhombus = 360°

The opposite angles of a rhombus are equal

The sum of the adjacent angles = 180° (The adjacent angles are supplementary)

Therefore, in the rhombus EFGH, we have;

Angle E, and angle H are supplementary angles

∴ Angle E + angle H = 180°

Substituting the values of angle E and angle H gives;

3·x + 5 + 4·x = 180°

7·x = 180° - 5 = 175°

x = 175°/7 = 25°

Angle H = 4·x

∴ Angle H = 4 × 25° = 100°

Angle H = 100°

Angle F = angle H = 100°

Therefore, angle F = 100°

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A rancher has 200 feet of fencing to enclose two adjacent corrals
Arturiano [62]

Answer:

a) Each corral should be 33⅓ ft long and 25 ft wide

b) The total enclosed area is 1666⅔ ft²

Step-by-step explanation:

I assume that the corrals have identical dimensions and are to be fenced as in the diagram below

Let x = one dimension of a corral

and y = the other dimension

 

(a) Dimensions to maximize the area

The total length of fencing used is:

4x + 3y = 200

4x = 200 – 3y

x = 50 - ¾y

The area of one corral is A = xy, so the area of the two corrals is

A = 2xy

Substitute the value of x

A = 2(50 - ¾y)y

A = 100 y – (³/₂)y²

This is the equation for a downward-pointing parabola:

A = (-³/₂)y² + 100y

a = -³/₂; b = 100; c = 0

The vertex (maximum) occurs at  

y = -b/(2a)  = 100 ÷ (2׳/₂) = 100 ÷ 3 = 33⅓ ft  

4x + 3y = 100

Substitute the value of y

4x + 3(33⅓) = 200

4x + 100 = 200

4x = 100  

x = 25 ft

Each corral should measure 33⅓ ft long and 25 ft wide.

Step 2. Calculate the total enclosed area

The enclosed area is 50 ft long and 33⅓ ft wide.

A = lw = 50 × 100/3 = 5000/3 = 1666⅔ ft²

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NEED HELP AND WILL GIVE BRAINLIEST!!!! Take a look below plz!!
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