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jek_recluse [69]
4 years ago
15

Match the reasons with the statements in the proof to prove DF = EF, given that DEF is a triangle by definition and angles 3 and

4 are equal.
Given:

DEF

3 = 4

Prove:

DF = EF

Mathematics
1 answer:
icang [17]4 years ago
6 0

Answer with explanation:

Given:

In Δ DEF, ∠3=∠4.

To prove:→ DE=E F

Proof:

1.  ∠3=∠4------[Given]

2. →∠1 and ∠ 3 are Supplementary to each other.

(a)⇒∠1 + ∠ 3=180°

 →∠2 and ∠ 4 are also Supplementary to each other.

(b)⇒∠2 + ∠ 4=180°

                                     --------------------[Exterior sides in opposite rays]

3. From 1 , a and b

 ⇒∠ 1 = ∠ 2-------[Two Angles Supplementary to equal Angles are equal to each other]

4. \Bar{DE}=\Bar{E F}

 If two angles of a Triangle are equal , then side opposite to these angles are equal.

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Answer: Proportional is two varying quantities are said to be in a relation of proportionality, if they are multiplicatively connected to a constant, that is, when either their ratio or their product yields a constant. The value of this constant is called the coefficient of proportionality or proportionality constant.

8 0
3 years ago
Solve. Will give brain thing
dezoksy [38]

Answer:

Step-by-step explanation:

-1/2 + (3/4 x 4/9)

-1/2 + (3/9)

= - 1/2 + (1/3)

= - 1/6

8 0
3 years ago
The wind speed near the center of a tornado is represented by the equation S=93logd+65, where d is the distance, in miles, that
dezoksy [38]

Answer:

Therefore,

Wind speed of a Tornado when it travels 4 miles is 121.0 miles per hour.

Step-by-step explanation:

Given:

The wind speed near the center of a tornado is represented by the equation,

S=93\log d+65

Where,

d =  the distance, in miles

S = the wind speed, in miles per hour.

To Find:

S = ? at d = 4 miles

Solution:

We have

S=93\log d+65        .........Given

Substitute d = 4 in above equation we get,

S=93\log 4 + 65=93\times 0.602 +65=120.99

Rounding the answer to the nearest tenth we have

S=121.0\ mph

Therefore,

Wind speed of a Tornado when it travels 4 miles is 121.0 miles per hour.

4 0
3 years ago
Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]
vampirchik [111]

Answer:1002

Step-by-step explanation:

\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )

and 1+\tan ^2\theta =\sec^2\theta

and \cos \theta =\frac{1}{\sec \theta }  

\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )

as \cos ^2\theta =\cos ^2(\pi -\theta )

Applying this we get

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]

every \thetathere exist \pi -\theta

such that \sin^2\theta +\cos^2\theta =1

therefore

\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002                          

6 0
3 years ago
Factor completely<br> 374 +48r2
irakobra [83]

Answer:

2(187 + 241r)

Step-by-step explanation:

3 0
3 years ago
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