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horrorfan [7]
1 year ago
15

What is the missing factor in the equation? Assume x * 0 and y* 0.

Mathematics
1 answer:
weeeeeb [17]1 year ago
4 0

Answer:

\sf b)  \quad \dfrac{y^3}{4x}

Explanation:

\sf \left(\dfrac{6x^2}{5y}\right)  \left(?}\right) = \left(\dfrac{3xy^2}{10}\right)

cross multiply variables

\sf ? = \left(\dfrac{3xy^2 (5y)}{10(6x^2)}\right)

distribute inside parenthesis

\sf ? = \left(\dfrac{15xy^3}{60x^2}\right)

remove parenthesis and simplify

\sf ? = \dfrac{y^3}{4x}

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Answer:

(x,y) =(-4,-3) --- Vertex

x = -4 --- Axis of symmetry

Step-by-step explanation:

Given

y = -6(x + 4)^2 - 3

Solving (a): The vertex

For an equation written in

y = a(x - h)^2 + k

The vertex is:

(x,y) = (h,k)

By comparison:

y = a(x - h)^2 + k  and y = -6(x + 4)^2 - 3

-h =4       k = -3

h =-4         k = -3  

So, the vertex is:

(x,y) =(-4,-3)

Solving (b): The axis of symmetry

For an equation written in

y = a(x - h)^2 + k

The axis of symmetry is:

x = h

In (a):

h =-4

So:

x = -4

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2 years ago
I need the answer quick before my timer goes out
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