Answer:
Step-by-step explanation:
hello :
16t²+120 = y
16t² = y -120 so : t² = (y-120)/16
if : y-120 ≥ 0 t = ±√((y-120)/16)
Answer:
By AA
ΔWXY ~ΔWVZ
Step-by-step explanation:
Here WXY is an isosceles triangle with legs WX & WY
So WX = WY
Hence ∠X = ∠Y
So ∠2= ∠3.
Now by angle sum property
∠1 + ∠2+∠3 = 180°
∠1+∠2+∠2=180°
2∠2 = 180° - ∠1 .......(1)
In triangle WVZ
WV = WZ
So ∠V = ∠Z
∠4 = ∠5
Once again by angle sum property
∠1 + ∠4 + ∠5=180°
∠1 + ∠4 + ∠4 = 180°
2∠4 = 180° - ∠1 ...(2)
From (1) & (2)
2∠2 = 2∠4
∠2=∠4
Now ∠W is common to both triangles
Hence by AA
ΔWXY ~ΔWVZ
Answer: x>- 7/3
Step-by-step explanation:
Solve for x by simplifying bot sides of the equation, then isolating the variable.