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joja [24]
3 years ago
9

A parabola is given by the equation y = x2 + 4x + 4.

Mathematics
1 answer:
notsponge [240]3 years ago
3 0

Answer:

the vertex of the parabola is:(-2,0)

the focus of the parabola is:(-2,\frac{1}{4})

the directrix of the parabola is:y=\frac{-1}{4}

Step-by-step explanation:

we know that for any general equation of the parabola of the type y=ax^{2} +b x+c the vertex of the parabola is given by (h,k)

where h=\frac{-b}{2 a} and k=\frac{4 a c-b^{2} }{4a}

therefore by the given data we have h=-2 and k=0

hence vertex=(-2,0)

the general equation of the parabola of the type 4 a (y-h)=(x-k)^2 ; the parabola symmetric around the y-axis has the focus from the centre i.e. the vertex (h,k) at a distance a as (h,k+a) and the directrix is given by y=k-a

so focus is (-2,\frac{1}{4} )

now the directrix of the parabola is y=\frac{-1}{4}.




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A pyramid-shaped vat has square cross-section and stands on its tip. The dimensions at the top are 2 m × 2 m, and the depth is 5
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the water level increases at a rate of 1.718 m/min when the depth is 4 m

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the volume of the pyramid is

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tg Ф = Side Length/ Height = L / H = x / h

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the volume of the water is

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in terms of time

v = Q*t

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Q*t = 1/3 (L/H)² h³

h³ = 3*(H/L)² *Q *t

h = ∛(3*(H/L)² *Q *t) = ∛(3*(H/L)² *Q) *∛t = k* ∛t , where k=∛(3*(H/L)² *Q)

h = k* ∛t

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dh/dt = 1/3*k* t^(-2/3)

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t = (h/k)³

dh/dt = 1/3*k* t^(-2/3) = 1/3*k* (h/k)³ ^(-2/3)  = 1/3*k* (h/k)^(-2) = 1/3 k³ / h²

=  1/3 (3*(H/L)²*Q) / h²  = (H/L)²*Q /h²

dh/dt= [H/(h*L)]²*Q

replacing values, when h=4m

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8 0
3 years ago
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