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OLEGan [10]
3 years ago
6

Solve similar triangles (advanced)

Mathematics
2 answers:
atroni [7]3 years ago
7 0

Answer:

<em>x = 24 </em>

Step-by-step explanation:

The corresponding sides of similar Δs are proportional.

ΔADE ~ ΔABC ⇒ \frac{AD}{AB} = \frac{DE}{BC} = \frac{AE}{AC}

AD = x + 8 ; DE = 8

AB = x ; BC = 6

\frac{x+8}{x} = \frac{8}{6} ⇔ 8x = 6(x + 8) ⇒ <em>x = 24</em>

tatiyna3 years ago
7 0

Answer:

<h3>x = 24</h3><h3></h3>

Step-by-step explanation:

<u>  6  </u>= <u>   x    </u>    

 8     x + 8

8x = 6x + 48

8x - 6x = 48

2x = 48

x = 48/2

x = 24

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Find the length of the third side of each triangle mark
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I'm going to use the Pythagoras theorem

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3 years ago
T=mv^2/L<br><br> Write an equation that shows the given formula solved for V
Effectus [21]

Answer:

<u>v = √(LT/m)</u>

Step-by-step explanation:

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Multiply both sides with L :

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3 0
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Axis of sym: x =
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Answer:

<h2>SEE BELOW</h2>

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • quadratic function
  • PEMDAS
<h3>let's solve:</h3>

vertex:(h,k)

therefore

vertex:(-1,4)

axis of symmetry:x=h

therefore

axis of symmetry:x=-1

  • to find the quadratic equation we need to figure out the vertex form of quadratic equation and then simply it to standard form i.e ax²+bx+c=0

vertex form of quadratic equation:

  • y=a(x-h)²+k

therefore

  • y=a(x-(-1))²+4
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it's to notice that we don't know what a is

therefore we have to figure it out

the graph crosses y-asix at (0,3) coordinates

so,

3=a(0+1)²+4

simplify parentheses:

3 = a(1 {)}^{2}  + 4

simplify exponent:

3 =  a + 4

therefore

a =  - 1

our vertex form of quadratic equation is

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let's simplify it to standard form

simplify square:

y =  - ( {x}^{2}  + 2x + 1)  + 4

simplify parentheses:

y =  -  {x}^{2}  - 2x - 1 + 4

simplify addition:

y =  -  {x}^{2}  - 2x + 3

therefore our answer is D)y=-x²-2x+3

the domain of the function

x\in \mathbb{R}

and the range of the function is

y\leqslant 4

zeroes of the function:

-  {x}^{2}  - 2x + 3 = 0

\sf divide \: both \: sides \: by \:  - 1

{x}^{2}  + 2x - 3 = 0

\implies \:  {x}^{2} +   3x  - x  +  3 = 0

factor out x and -1 respectively:

\sf \implies \: x(x + 3)   - 1(x  + 3 )= 0

group:

\implies \: (x - 1)(x + 3) = 0

therefore

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