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Vadim26 [7]
3 years ago
14

write an equation for a function that has a graph with the shape of y=x^2, but upsidedown and shifted right 6 units and up 4 uni

ts. ​
Mathematics
1 answer:
irina1246 [14]3 years ago
5 0

Answer:

y - 4 = -(x - h)^2

Step-by-step explanation:

We begin with the stock function y = x^2.  Its graph is a parabola with vertex at (0, 0) and opening up.  The more general vertex form of the graph is

y - k = a(x - h)^2, where the point (h, k) is the vertex and 'a' determines the direction of opening as well as by how much the parabola is stretched vertically.

In this problem the vertex is (6, 4); that is, the vertex is shifted to the right by 6 units and upward from there by 4 units.  The coefficient 'a' is -1, indicating that the graph opens downward.

So we have the equation y - 4 = -(x - h)^2.  Without more information we cannot tell whether the graph is stretched vertically or not.

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Answer:

Step-by-step explanation:

hello :

cot x + 2cot x  sin x = 0        0≤ x≤ 180

cotx(1+2sinx) =0    means : cotx=0   or  1+2sinx=0

1)  cotx=0   so :  cosx=0  and  sinx ≠ 0   because cotx = cosx/sinx

when :  0≤ x≤ 180  : x=90°

2) 1+2sinx=0  means : sinx = -1/2

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The equation of the given surface is,

z=g(x,y)=xy

Solving the partial derivatives,

∂g∂x=y,∂g∂y=x

Substituting to the formula

S=∬√1+( ∂g∂x)2+(∂g∂y)2dA

Thus,

S=∬√1+(y)2+(x)2dAS=∬√1+x2+y2dA

The region in the XY-plane is defined by the intervals  0≤θ≤2π,0≤r≤4

Converting the integral into polar coordinates,

S=∫2π0∫40√1+r2rdrdθ

Integrating with respect to r

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Integrating with respect to θ

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