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LenKa [72]
3 years ago
14

Given: ΔABC ≅ ΔEFD What is the length of Segment line FE rounded to the nearest tenth?

Mathematics
2 answers:
disa [49]3 years ago
8 0
Line segment FE is equals to Line segment BA or FE=BA.

B =  (2,4) A = (0,2) 

First we have to subtract the x and y coordinates. 

x = |2 - 0|
x = 2

y = |4-2|
y = 2

Then, we are going to use the Pythagorean Theorem a^2<span> + b^</span>2<span> = c^</span><span>2.

2^2 + 2^2 = 8

Get the square root of 8. The answer will be 2.83. Rounded off to the nearest tenth, the answer will be 2.80

I hope this helped you.

</span>
 
Tpy6a [65]3 years ago
4 0

The answer is A. 2.8. Explained: I used the distance formula (or the top part of Pythagorean theorem) of d=square root of (x2-x1)+(y2-y1). Using points A (0,2) as (x1,y1) and point B (2,4) as (x2,y2) to put inside the distance formula. d=square root of (2-0)^2+(4-2)^2= square root of (2)^2+(2)^2= square root of (4+4)= square root of 8 which equals 2.8. Since triangle ABC is congruent to triangle EFD then since side AB equals 2.8 then side FE equals 2.8.

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3 years ago
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2 years ago
Find the angle between the given vectors to the nearest tenth of a degree. u = &lt;-5, -4&gt;, v = &lt;-4, -3&gt; (1 point)
sesenic [268]

Angle between u = -5i-4j , v=-4i-3j is x =0° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3>  or , u = -5i-4j , v=-4i-3j

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

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Now , u.v = (-5i-4j)(-4i-3j)

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Now , Modulus of any vector  r = xi+yj is |r| = \sqrt{x^{2}+y^{2}} So ,

|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5

Putting all these values in equation cosx =\frac{u.v}{|u|(|v|)} we get:

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⇒ x =cos^{-1}(1)                 { cos0 = 1  }

⇒ x =0°

Therefore , Angle between u = -5i-4j , v=-4i-3j is x =0° .

8 0
3 years ago
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the answer is x=(1/28)y^2

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Justification:

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3 years ago
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