Answer:// Solve equation [1] for the variable y
[1] y = 2x - 3
// Plug this in for variable y in equation [2]
[2] -2•(2x-3) + 2x = 2
[2] - 2x = -4
// Solve equation [2] for the variable x
[2] 2x = 4
[2] x = 2
// By now we know this much :// Solve equation [1] for the variable y
[1] y = 2x - 3
// Plug this in for variable y in equation [2]
[2] -2•(2x-3) + 2x = 2
[2] - 2x = -4
// Solve equation [2] for the variable x
[2] 2x = 4
[2] x = 2
// By now we know this much :
y = 2x-3
x = 2
// Use the x value to solve for y
y = 2(2)-3 = 1
y = 2x-3
x = 2
// Use the x value to solve for y
y = 2(2)-3 = 1
Step-by-step explanation:
Yes, the bisecting line will also bisect the side
Answer:
.
Step-by-step explanation:
Dont know if this helps but heres something.
A = πr²/2 = 1938.900992 * π [yd²] ≈ 4.7 [yd²]
Use this if you want to
https://www.omnicalculator.com/math/semicircle-area
For the first one, you did good. I will just suggest a couple things.
Statement Reason
JK ≅ LM Given
<JKM ≅ < LMK Given (You did both of these steps so well done.)
MK ≅ MK Reflexive Property (Your angle pair is congruent but isn't one of the interior angle of the triangles you are trying to prove.)
ΔJMK ≅ ΔLKM SAS
Problem 2: (You also have a lot of great stuff here.)
Statement Reason
DE ║ FG Given
DE ≅ FG Given
<DEF≅<FGH Given
<EDF≅<GFH Corresponding Angles (You don't need to know that F is the midpoint but you got corresponding angle pair which is correct.)
ΔEDF≅ΔGFH ASA
<DFE≅<FHG CPCTC
