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3 years ago

How do you find an equivalent ratio?

1 answer:

5 0

It can be written two ways; for example, 1:10 or 1/10. Equivalent ratios are two ratios that express the same relationship between numbers. You can create equivalent ratios by multiplying or dividing both the numerator and denominator of a given ratio by the same number.

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OmZC > mZE > m2D

matrenka [14]

Answer:

Step-by-step explanation: Answer to ol in your answer.) What are mZA, mZB, and mZC? mZA= (Round to one decimal place as needed. Do not include the degree. I hope this either gives you a quick hint and a tip. Thank you and have a good day.

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3 years ago

Find the mean for the following sets of numbers.14,15,20,21

Jet001 [13]

The mean of those sets of numbers are 17.5

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3 years ago

Read 2 more answers

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.