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kati45 [8]
3 years ago
13

What is this the relationship between the legs and the hypotenuse of a 45 - 45 - 90 triangle?

Mathematics
2 answers:
Studentka2010 [4]3 years ago
6 0
The length of the hypotenuse is (the length of each leg) times (the square root of 2).
Elena L [17]3 years ago
6 0

Answer:

Step-by-step explanation:

The length of the hypotenuse and the legs of a 45 45 90 triangle are no different from any right triangle. It's simply:

a^2+b^2=c^2 where a and b are the legs and c is the hypotenuse.

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5.7-1.5=4.2

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Can someone help me with this<br> S=?
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In order to find the value of s, we must isolate the s variable in this equation, so the first step should be subtracting 3 5/6 from both sides of this equation. In order to subtract, we multiply each denominator to get 42, and multiply the numerator by the same amount, and subtract the number from both sides:

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If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle
stira [4]

Answer:

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

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

Given points end  Points are p(-3,-2) and q( 1,6)

<em>The distance of two points formula</em>

P Q = \sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2}   }

P Q = \sqrt{1 - (-3)^{2} + ((6 -(-2))^{2}   }

P Q = \sqrt{16+64} = \sqrt{80}

<em>The diameter 'd' = 2 r</em>

                       2 r = √80

                            = \sqrt{16 X 5}

                           = 4 \sqrt{5}

                     <em> r =  2√5</em>

<em>Mid-point of two end points </em>

<em>                        </em>(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )<em></em>

<em>                                               = (-1 ,2)</em>

<em>Mid-point of two end points = center of the circle</em>

<em>                                     (h,k) = (-1 , 2)</em>

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

<em>          (x -(-1) )² +(y-(2))² = (2(√5))²</em>

<em>         x² + 2 x + 1 + y² - 4 y + 4 = 20</em>

<em>          x² + 2 x + y² - 4 y  = 20 -5</em>

<em>           x² + 2 x + y² - 4 y  = 15</em>

<u><em>Final answer</em></u><em>:-</em>

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<em>                  </em>

6 0
4 years ago
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