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Ad libitum [116K]
3 years ago
12

(c + 3) - 2c - (1 - 3c) = 2 what is this answer to this problem?

Mathematics
1 answer:
Eddi Din [679]3 years ago
6 0
(c+3)-2c-(1-3c)=2
c+3-2c-1+3c=2
2c+2=2
2c=0
c=0
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Small plane circles the airport the angle of depression would be 180
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Polygon F has an area of 36 square units. Aimar drew a scaled version of Polygon F and labeled it Polygon G. Polygon G has an ar
Free_Kalibri [48]

Answer:

1/3

Step-by-step explanation:

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

Each side of Polygon FFF was multiplied by a certain value, known as the scale factor , to result in an area that is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF.

[Show me an example of how scale factor affects area]

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction

 

 

\begin{aligned} A &= \left(l\times\dfrac1{10}\right)\times\left(w\times\dfrac1{10}\right) \\ \\ A&= l\times w\times\dfrac1{10}\times\dfrac1{10} \\ \\ A&= lw \times \left(\dfrac1{10}\right)^2\end{aligned}  

 

 

 

 

 

 

 

 

\dfrac1{10}  

start fraction, 1, divided by, 10, end fraction\left(\dfrac1{10}\right)^2  

 

left parenthesis, start fraction, 1, divided by, 10, end fraction, right parenthesis, start superscript, 2, end superscript

Hint #22 / 3

The area of a polygon created with a scale factor of \dfrac1x  

x

1

​  start fraction, 1, divided by, x, end fraction has \left(\dfrac1{x}\right)^2(  

x

1

​  )  

2

left parenthesis, start fraction, 1, divided by, x, end fraction, right parenthesis, start superscript, 2, end superscript the area of the original polygon:

\left(\text{scale factor}\right)^2=\text{fraction of the area the scale copy has}(scale factor)  

2

=fraction of the area the scale copy hasleft parenthesis, s, c, a, l, e, space, f, a, c, t, o, r, right parenthesis, start superscript, 2, end superscript, equals, f, r, a, c, t, i, o, n, space, o, f, space, t, h, e, space, a, r, e, a, space, t, h, e, space, s, c, a, l, e, space, c, o, p, y, space, h, a, s

The area of Polygon GGG is \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction the area of Polygon FFF. Let's substitute \dfrac19  

9

1

​  start fraction, 1, divided by, 9, end fraction into the equation to find the scale factor.

\left(\dfrac1{?}\right)^2=\dfrac19(  

?

1

​  )  

2

=  

9

1

​  left parenthesis, start fraction, 1, divided by, question mark, end fraction, right parenthesis, start superscript, 2, end superscript, equals, start fraction, 1, divided by, 9, end fraction

The scale factor is \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction.

Hint #33 / 3

Aimar used a scale factor of \dfrac13  

3

1

​  start fraction, 1, divided by, 3, end fraction to go from Polygon FFF to Polygon GGG.

4 0
3 years ago
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Find the tangents of the acute angles in the right triangle. Write each answer as a fraction. WILL MAKE BRAINLIEST!!
patriot [66]

Answers:

  • tan(G) = 2
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=======================================

Explanation:

Recall that tangent is the ratio of opposite over adjacent

tan(angle) = opposite/adjacent

So for reference angle G, we say,

tan(G) = JH/GJ = 2/1 = 2

We'll treat tan(H) in a similar fashion, but the opposite and adjacent sides swap roles. That means we'll apply the reciprocal to the result above to get 1/2 for tan(H)

-----------

So we have this interesting property where

tan(G)*tan(H) = 2*(1/2) = 1

In general,

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-----------

Side note: The side sqrt(5) isn't used at all.

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