Answer:
Step-by-step explanation:
1S. JM bisects ∠ KJL and ∠JMK ≅ ∠JML
1R. Given
2S. ∠KJM ≅ ∠LJM
2R. Angle bisectors form ≅ ∠s
3S. JM≅JM
3R. Reflexive Propriety ( segment is ≅ with itself)
4S. ΔJMK ≅ΔJML
4R. ASA theorem of congruency
5S. JK≅JL
5R. Coresponding parts of congruent Δs are congruent (CPCTC)
Answer:
ok boomer
Step-by-step explanation:
because i said so also ban me if you want i dont care little 1+1
Answer:
Step-by-step explanation:
Perimeter is equal to twice the length plus twice the width
Perimeter = 2L + 2W factoring out the 2
Perimeter = 2(L + W) L = (8x + 5 - 3x) W = (4x - 1)
first I would simplify L collecting the terms of x
L = (5x + 5)
plug in to L and W
Perimeter = 2 (( 5x + 5 ) + ( 4x - 1 )) L = (8x + 5 - 3x) W = (4x - 1)
Perimeter = 2 ( 5x + 4x + 5 -1 ) collect like terms
Perimeter = 2 ( 9x +4) 2 times the 9 and 4
Perimeter = 18x + 8 SOLVE for x
Perimeter - 8 = 18x + 8 -8
Perimeter - 8 = 18x
(Perimeter - 8) / 18 = 18x / 18
(Perimeter - 8) / 18 = x
remember the Perimeter equals 88
(88 - 8) / 18 = x units
4.444444444 units = x
This seems like an odd number so let's it
Perimeter = 2 (( 5x + 5 ) + ( 4x - 1 ))
Perimeter = 2 (( 5(4.444) + 5 ) + ( 4(4.444) - 1 )) = 87.9952
close enough with the rounding off error
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