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Basile [38]
2 years ago
8

In quadrilateral HJKL shown below, point J can be mapped to point K by a 46 counterclockwise rotation about point H, and point K

can be mapped to point L by a 72 counterclockwise rotation about point H.
Explain why HJ must be congruent to HL.
Determine the measure of HJL

Mathematics
1 answer:
vladimir2022 [97]2 years ago
3 0

Answer:

Just kidding I willa asnwer for real

Step-by-step explanation:

They are  Hj and Hl because it is part a the quadrilateral, rhombus

Hjl is 46

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Point-Slope form:  y+1=-\frac{6}{15}(x+5)\\

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Step-by-step explanation:

The standard form is given by: Ax+By=C

The point-slope form is given by: y-y_1=m(x-x_1)


  • <em>Let's call (-5,-1) as point (x_1,y_1), and</em>

<em>point (10,-7) as point (x_2,y_2)</em>

  • <em>We also know that the slope m is given by m=\frac{y_2-y_1}{x_2-x_1}</em>

Let's find the slope:

m=\frac{-7-(-1)}{10-(-5)}=-\frac{6}{15}


1. So, point slope form can be written as:

y-(-1)=-\frac{6}{15}(x-(-5))\\y+1=-\frac{6}{15}(x+5)\\


2. Rearranging this equation and bringing all x's and y's to one side and the number to another side will give us the standard form. So:

y+1=-\frac{6}{15}x-2\\y+\frac{6}{15}x=-2-1\\y+\frac{6}{15}x=-3

<em>But A, B, and C (standard form) needs to be integers, so to get rid of the denominator , we multiply the whole equation by 15. So we have:</em>

y+\frac{6}{15}x=-3\\15y+6x=-45

<em>We can factor out a 3, so it becomes:</em>

15y+6x=-45\\5y+2x=-15

So standard form is 2x+5y=-15

8 0
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