y = x^2 + 2x...eqn 1
y = 3x + 20...eqn 2
subst for y in eqn 1...
=> x^2 +2x = 3x +20
=> x^2 - x - 20 =0
=> (x-5) (x+4) =0
=> x = 5 or -4
for x =5, y = 35 (sub for x in eqn 1 or 2)
for x = -4, y = 8 (sub for x in eqn 1 or 2)
Step-by-step explanation:
by Pythagoras theorem:
perpendicular² + base²= hypotenuse²
(1/√2)²+ (1/√3)² = H²
1/2+1/3= H²
5/6 = H²
√(5/6 )= H
√5/√6 = H
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Answer:
have you tried using math-way just with no -
Step-by-step explanation:
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