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Lyrx [107]
3 years ago
14

Need Help

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0

Answer:

2\sqrt{18}+6 \ in

Step-by-step explanation:

As ABC is an isosceles triangle, the segment CD cuts to AB in two equal parts. Then, AD=DB=3 in. Now, using Pitagoras Theorem we have that:

CD^2+DB^2 = BC^2

3^2+3^2 = BC^2

BC = \sqrt{3^2+3^2}

BC = \sqrt{9+9}

BC = \sqrt{9+9}

BC = \sqrt{18}=AC.

Now, the perimeter is the sum of the three sides of the triangle, then

perimeter = AC+BC+AB = \sqrt{18}+\sqrt{18}+6 = 2\sqrt{18}+6 \ in.

anzhelika [568]3 years ago
5 0
2x+6=y
triangle ADC=x+3+3=y
triangle CBD=x+3+3=y
this is not possible until u give perimeter 
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Soal soal kesebangunan segitiga
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Answer:

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Step-by-step explanation:

5 0
3 years ago
Rewrite following sets in roster form or listing form.
pshichka [43]

Answer:

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8 0
3 years ago
A zip-line at a water park runs from the top of a wall, down to the ground on the other side of a pool. The top of the zip cord
Contact [7]

Answer:

498 ft

Step-by-step explanation:

The zip-line, wall and end of the zip-line to wall form a right-angled triangle.

Since the zip-line which is 500 ft long represents the hypotenuse side, the top of the zip cord to the ground represents one side of the triangle which is 45 ft long and the third side is represented by the horizontal distance between the wall and the zip-line.

By Pythagoras' theorem,

L² = h² + d² where L = length of zip-line = 500 ft, h = height of top of zip cord from ground = 45 ft and d = horizontal distance between the wall and the zip-line.

So, d² = L² - h²

d² = (500 ft)² - (45 ft)²

d² = 250,000 ft² - 2025 ft²

d² = 247,975 ft²

d = √247,975 ft²

d = 497.97 ft

d ≅ 498 ft

7 0
3 years ago
Given quadrilateral RSTU, determine if each pair of sides (if any) are parallel and which are perpendicular for the coordinates
WINSTONCH [101]
Check the picture below.

so, clearly UR is not parallel to ST, but is likely that UT is parallel to RS.

keeping in mind that parallel lines, have the same exact slope, let's check the slope for UT as well as RS, if they are the same, then indeed both are parallel,

\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&R&(~ 1 &,& -3~) 
%  (c,d)
&S&(~ 4 &,& -1~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-(-3)}{4-1}\implies \cfrac{-1+3}{4-1}\implies \cfrac{2}{3}\\\\
-------------------------------

\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&U&(~ -4 &,& -2~) 
%  (c,d)
&T&(~ 2 &,& 2~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-(-2)}{2-(-4)}\implies \cfrac{2+2}{2+4}\implies \cfrac{4}{6}\implies \cfrac{2}{3}

there you have it, notice the slopes of each.

from the picture, clearly there are no right-angles at vertices U and R, however at S and T, it looks like, but we dunno.

well, we know UT || RS, now if ST ⟂ RS, then ST will have a negative reciprocal slope  to RS, namely whatever the slope of RS is, ST will be the negative reciprocal of that, so let's check what is the slope of ST then,
 
\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{2}{3}\\\\
negative\implies  -\cfrac{2}{ 3}\qquad reciprocal\implies \boxed{- \cfrac{ 3}{2}}\\\\
-------------------------------\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&S&(~ 4 &,& -1~) 
%  (c,d)
&T&(~ 2 &,& 2~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{2-(-1)}{2-4}\implies \cfrac{2+1}{2-4}\implies \boxed{-\cfrac{3}{2}}

 and if ST ⟂ RS, then ST ⟂ UT.

8 0
3 years ago
Please answer both I’ll give brainliest
Readme [11.4K]

Answer:

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