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sertanlavr [38]
3 years ago
10

if gabby wants to make a regular octagon with a side length of 20 inches of water , how much wire does she need?

Mathematics
1 answer:
EleoNora [17]3 years ago
4 0

Answer:

160 inches

Step-by-step explanation:

An octagon has 8 sides;

20*8=160

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Answer: 7/8

Step-by-step explanation:

You can change all the fractions to have equivalent denominators.

The Least Common Denominator of 1 1/4 and 3/8 is 8. Now we can change the fractions.

1 1/4 = 1 2/8 = 10/8.

3/8 = 3/8.

Now we can subtract. Subtract the numerators of the first fraction from the numerators of the second fraction, and the denominators oft he first fraction to the denominators of the second fraction.

10/8 - 3/8 = 7/8.

Therefore, 1 1/4 - 3/8 = 7/8; the fourth answer.

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2a + 4b + c =5 <br> x =-2 <br><br> a - 4b = -6<br> y=1
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There is no solution ,<span>a+c=-10;b-c=15;a-2b+c=-5 </span>No solution System of Linear Equations entered : [1] 2a+c=-10 [2] b-c=15 [3] a-2b+c=-5 Equations Simplified or Rearranged :<span><span>  [1] 2a + c = -10 </span><span>  [2] - c + b = 15 </span><span>  [3] a + c - 2b = -5 </span></span>Solve by Substitution :

// Solve equation [3] for the variable  c  
 

<span> [3] c = -a + 2b - 5 </span>

// Plug this in for variable  c  in equation [1]

<span><span>  [1] 2a + (-a +2?-5) = -10 </span><span>  [1] a = -5 </span></span>

// Plug this in for variable  c  in equation [2]

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// Solve equation [2] for the variable  ?  
 

<span> [2] ? = b + 10 </span>

// Plug this in for variable  ?  in equation [1]

<span><span>  [1] (? +10) = -5 </span><span>  [1] 0 = -15 => NO solution </span></span><span>No solution</span>
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Can someone help me with this question on Prodigy? ​
lions [1.4K]

Answer:

  8 +(3/8)√53 in² ≈ 10.73 in²

Step-by-step explanation:

Given the net of a triangular pyramid with some of the dimensions filled in, you want to find the total surface area.

<h3>Triangle base</h3>

The triangle bases identified by dashed lines will have a length equal to the hypotenuse of the right triangles with legs shown as solid lines. The legs of each of those right triangles are ...

  a = (3 in)/2 = 1.5 in

  b = 2 in . . . . . . shown as the altitude of the triangle

Then the hypotenuse is found using the Pythagorean theorem:

  c² = a² +b²

  c² = 1.5² +2² = 2.25 +4 = 6.25

  c = √6.25 = 2.5

The dashed lines are 2.5 inches long.

<h3>Triangle altitude</h3>

The altitude from the solid horizontal line to the vertex at the bottom of the figure can be found using the fact that all of the outside edge lengths of the net are the same length. That edge length is found as the length of the hypotenuse of the right triangles in the left- and right-sides of the upper portion of the net. Each of those has a leg that is (2.5 in)/2 = 1.25 in and a leg marked as 2 in.

  c² = a² +b²

  c² = 1.25² +2² = 1.5625 +4 = 5.5625

  c = (√89)/4 ≈ 2.358 . . . in

The unmarked altitude of the bottom triangle is then ...

  b² = c² -a²

  b² = 89/16 -1.5² = 53/16

  b = (√53)/4 ≈ 1.820 . . . in

<h3>Surface area</h3>

The surface area of the figure is the sum of the areas of the four triangles that make up the net. Each triangle has an area given by the formula ...

  A = 1/2bh

The left and right triangles have b=2.5, h=2, so they each have an area of ...

  A = 1/2(2.5)(2) = 2.5 . . . . in²

The center triangle has dimensions of b=3, h=2, so an area of ...

  A = 1/2(3)(2) = 3 . . . . in²

The bottom triangle has dimensions of b=3, h=(√53)/4, so an area of ...

  A = 1/2(3)(√53/4) = (3/8)√53 ≈ 2.730 . . . . in²

The total surface area is the sum of the areas of these triangles, so is ...

  A = 2.5 in² +2.5 in² +3 in² +2.73 in² = 10.73 in²

The surface area of the triangular pyramid is (64+3√53)/8 ≈ 10.73 in².

__

<em>Additional comment</em>

Often we work with pyramids that are rotationally symmetrical about a vertical line through the peak. This one is not. The altitude of the bottom triangle in the net is less than the altitude of the other triangles. This short face of the pyramid will tend to be more vertical than the other two lateral faces.

4 0
1 year ago
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