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lina2011 [118]
3 years ago
9

Can you please solve the following equation?

Mathematics
1 answer:
AlladinOne [14]3 years ago
7 0

Answer:

x=3.903

Step-by-step explanation:

2^{x-3}.5^{x-1}=200\\2^x.2^{-3}.5^x.5^{-1}=200\\\frac{2^x}{2^3}.\frac{5^x}{5^1}=200\\\frac{2^x.5^x}{2^3.5}=200\\\frac{(2.5)^x}{8.5}=200\\10^x=200 \times 40\\10^x=8000\\log 10^x=log8000\\x=\frac {log8000}{log10}=log8000=log 8\times log1000=log 8+log 1000=log2^3+log10^3=3log2+3log10=3log2+3 \approx0.903+3 \approx3.903

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The boundary of the lawn in front of a building is represented by the parabola . The parabola is represented on the coordinate p
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The boundary of the lawn in front of a building is represented by the parabola

y = (x^2) /16 + x - 2

And you have three questions which require to find the focus, the vertex and the directrix of the parabola.

Note that it is a regular parabola (its symmetry axis is paralell to the y-axis).

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It is a point on the symmetry axis => x = the x-component of the vertex) at a distance equal to the distance between the directrix and the vertex).

In a regular parabola, the y - coordinate of the focus is p units from the y-coordinate of the focus, and p is equal to 1/(4a), where a is the coefficient that appears in this form of the parabola's equation: y = a(x - h)^2 + k (this is called the vertex form)

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What we need is to complete a square. You can follow these steps.

1) Extract common factor 1/16 => (1/16) [ (x^2) + 16x - 32]

2) Add (and subtract) the square of the half value of the coefficent ot the term on x =>

16/2 = 8 => add and subtract 8^2 => (1/16) [ (x^2) + 16 x + 8^2 - 32 - 8^2]

3) The three first terms inside the square brackets are a perfect square trinomial: =>

(1/16) [ (x+8)^2 - 32 - 64] = (1/16) [ (x+8)^2  - 96] =>

(1/16) [(x+8)^2 ] - 96/16 =>

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Which is now in the form a(x - h)^2 + k,
where:

a = 1/16 , h = - 8, and k = -6

(h,k) is the vertex: h is the x-coordinate of the vertex, and k is the y-coordinate of the vertex.

=> a = 1/16 => p =1/4a = 16/4 = 4
 
y-componente of the focus = -6 + 4 = -2

x-component of the focus = h = - 8

=> focus = (-8, -2)

2) Vertex

We found it above, vertex = (h,k) = (-8,-6)


3) Directrix

It is the line y = p units below the vertex = > y = -6 - 4 = -10

y = -10   

 


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