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Softa [21]
4 years ago
7

Name the special segment. AC

Mathematics
1 answer:
katovenus [111]4 years ago
7 0
Pretty sure it is altitude sorry if im wrong
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Help please I am very confused
Helga [31]

The sides of the triangle occur in a ratio of 4 : 7 : 2, so if <em>x</em> is some positive number, then we can write each side's length in terms of <em>x</em> as 4<em>x</em>, 7<em>x</em>, and 2<em>x</em>.

The perimeter is 299 yd, so

4<em>x</em> + 7<em>x</em> + 2<em>x</em> = 299 yd

13<em>x</em> = 299 yd

<em>x</em> = (299 yd) / 13

<em>x</em> = 23 yd

Then the sides of the triangle have lengths of

4<em>x</em> = 4 • 23 yd = 104 yd

7<em>x</em> = 7 • 23 yd = 161 yd

2<em>x</em> = 2 • 23 yd = 46 yd

"Median" here refers to the side length between the shortest and longest sides, so the answer would be 104 yd.

3 0
3 years ago
a factory makes parts for toys in different quantities,as shown in the table. how much would 11 part cost?
Paraphin [41]

Answer can you show use the graph then i would be able to answer it


5 0
3 years ago
The figure below shows a shaded region and a nonshaded region. Angles in the figure that appear to be right angles are right ang
Ulleksa [173]
<span>The "grey part" is 87 the rest is the "shaded part" together they add up to the big red rectangle = 128</span>
3 0
3 years ago
Read 2 more answers
Please answer thanks
MA_775_DIABLO [31]
7. 21.39
8. 6
those are the only two i know

8 0
3 years ago
Analyze the diagram below and answer the question that follows
laiz [17]

Answer:

C. 40.2°

Step-by-step explanation:

Cosine rule (real handy to remember): c² = a² + b² - 2·a·b·cos(γ)

If you don't know this yet, look it up but in short: c, a and b are the lengths of the sides of the triangle, the angle opposite side a is called α, for b it is β and for c it is γ. That's the convention I've always used anyway, you can call them whatever of course. Anyhow:

c² = a² + b² - 2·a·b·cos(γ)

⇒ |AC|² = |AB|²+|BC|²-2·|AB|·|BC|·cos(∠B)

⇒ |AC|²-|AB|²-|BC|² = -2·|AB|·|BC|·cos(∠B)

⇒ ( |AC|²-|AB|²-|BC|² ) / ( -2·|AB|·|BC| ) = cos(∠B)

⇒ ∠B = arccos( ( |AC|²-|AB|²-|BC|² ) / ( -2·|AB|·|BC| ) )

         = arccos( ( 11²-16²-16² ) / ( -2·16·16 ) )

         = 40.21101958°

         ≈ 40.2°

6 0
2 years ago
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