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murzikaleks [220]
3 years ago
10

Identify each figure in as many ways as possible

Mathematics
1 answer:
sattari [20]3 years ago
5 0
Parallelogram
- quadrilateral (four sided figure)
- oposite sides parallel
<span>Rectangle
</span>- quadrilateral
- equiangular
<span>- right angle (90</span>° angle)<span>
Isosceles Trapezoid
</span>- quadrilateral
- pair of parallel sides
- called "trampezium" in the UK
<span>Square
</span>- regular quadrilateral
- equiangular
- right angles (90°)
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100 POINTS!!!!!!!!!! PLEASE HELP NEED THIS DONE AS SOON AS POSSIBLE ASAP
Llana [10]

Answer:

20.28

Step-by-step explanation:

Triangles; 4

Area of 1 Triangle; 2

Rectangles; 8

Semi Circle; 6.28

Area in total; 20.28

Hope This Helps! :)

4 0
2 years ago
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PLS ANSWER AS SOON AS POSSIBLE
Crank
I’m think it’s true! Good luck!
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3 years ago
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What is 5 1/2 * 2 1/3?
storchak [24]

The solution of the given expression will be 12\dfrac{5}{6}

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition,substraction, multiplication and division.

The given expression will be calculated as:-

= 5\dfrac{1}{2}\times 2\dfrac{1}{3}\\\\\\=\dfrac{11}{2}\times \dfrac{7}{3}\\\\\\=\dfrac{77}{6}\\\\\\=12\dfrac{5}{6}

Hence the solution of the given expression will be 12\dfrac{5}{6}

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5 0
2 years ago
What is the least common denominator of 1/7,2/5 and 2/3?
Maurinko [17]

Answer:

105 is the least common denominator

Step-by-step explanation:

Since 7, 5 and 3 are all prime.

LCM would be 7×3×5 = 105

5 0
4 years ago
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I'm have some issues with the problem shown in the screenshot and would love some help.
KiRa [710]

The dimensions and volume of the largest box formed by the 18 in. by 35 in. cardboard are;

  • Width ≈ 8.89 in., length ≈ 24.89 in., height ≈ 4.55 in.

  • Maximum volume of the box is approximately 1048.6 in.³

<h3>How can the dimensions and volume of the box be calculated?</h3>

The given dimensions of the cardboard are;

Width = 18 inches

Length = 35 inches

Let <em>x </em>represent the side lengths of the cut squares, we have;

Width of the box formed = 18 - 2•x

Length of the box = 35 - 2•x

Height of the box = x

Volume, <em>V</em>, of the box is therefore;

V = (18 - 2•x) × (35 - 2•x) × x = 4•x³ - 106•x² + 630•x

By differentiation, at the extreme locations, we have;

\frac{d V }{dx}  =  \frac{d( 4 \cdot \:  {x}^{3}  - 106 \cdot \:  {x}^{2}  + 630\cdot \:  {x} )  }{dx} = 0

Which gives;

\frac{d V }{dx}  =12\cdot \:  {x}^{2}  -  212\cdot \:  {x} + 630 = 0

6•x² - 106•x + 315 = 0

x  =  \frac{ - 6 \pm \sqrt{106 ^2 - 4 \times 6 \times 315} }{2 \times 6}

Therefore;

x ≈ 4.55, or x ≈ -5.55

When x ≈ 4.55, we have;

V = 4•x³ - 106•x² + 630•x

Which gives;

V ≈ 1048.6

When x ≈ -5.55, we have;

V ≈ -7450.8

The dimensions of the box that gives the maximum volume are therefore;

  • Width ≈ 18 - 2×4.55 in. = 8.89 in.

  • Length of the box ≈ 35 - 2×4.55 in. = 24.89 in.

  • Height = x ≈ 4.55 in.

  • The maximum volume of the box, <em>V </em><em> </em>≈ 1048.6 in.³

Learn more about differentiation and integration here:

brainly.com/question/13058734

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7 0
1 year ago
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