The answer to 11 pleasee?
1 answer:
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We know that
the equation of a vertical parabola in vertex form
y=a*(x-h)²+k
(h,k)------> (0,5)
y=a*(x-0)²+5
y=a*x²+5
substitute the point (2,9) in the equation
9=a*(2)²+5------> 9=4*a+5-------> 4*a=9-5-----> 4*a=4-----> a=1
the equation of the vertical parabola is
y=x²+5
the equation of a horizontal parabola in vertex form
x=a*(y-k)²+h
(h,k)------> (0,5)
x=a*(y-5)²+0
x=a*(y-5)²
substitute the point (2,9) in the equation
2=a*(9-5)²------> 2=16*a------> a=1/8
the equation of the horizontal parabola is
x=(1/8)*(y-5)²
the answer is
the equation of the vertical parabola is y=x²+5
the equation of the horizontal parabola is x=(1/8)*(y-5)²
see the attached figure
the equation of a vertical parabola in vertex form
y=a*(x-h)²+k
(h,k)------> (0,5)
y=a*(x-0)²+5
y=a*x²+5
substitute the point (2,9) in the equation
9=a*(2)²+5------> 9=4*a+5-------> 4*a=9-5-----> 4*a=4-----> a=1
the equation of the vertical parabola is
y=x²+5
the equation of a horizontal parabola in vertex form
x=a*(y-k)²+h
(h,k)------> (0,5)
x=a*(y-5)²+0
x=a*(y-5)²
substitute the point (2,9) in the equation
2=a*(9-5)²------> 2=16*a------> a=1/8
the equation of the horizontal parabola is
x=(1/8)*(y-5)²
the answer is
the equation of the vertical parabola is y=x²+5
the equation of the horizontal parabola is x=(1/8)*(y-5)²
see the attached figure
Step-by-step explanation:
Hey there!
By looking through figure, l and m are parallel lines and a transversal line passes through the lines.
Now,
7x + 12° = 12x - 28°( alternate angles are equal)
12°+28° = 12x - 7x
40° = 5x
Therefore, x = 8°
Now,
12x - 28° + 9y - 77 = 180° ( being linear pair)
12×8° - 28° + 9y -77° = 180°
96° - 28° + 9y - 77° = 180°
-9 + 9y = 180°
9y = 180° + 9°
y = 189°/9
Therefore, y = 21°
<u>There</u><u>fore</u><u>,</u><u> </u><u>X </u><u>=</u><u> </u><u>8</u><u>°</u><u> </u><u>and</u><u> </u><u>y</u><u>=</u><u> </u><u>2</u><u>1</u><u>°</u><u> </u><u>.</u>
<em><u>Hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>