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ale4655 [162]
3 years ago
12

Find the vertex and length of the latus rectum for the parabola. y=-1/2(x-2)^2+8

Mathematics
1 answer:
victus00 [196]3 years ago
4 0
Y = -(1/2)(x-2)² +8
Re write it in standard form:
(y-8) = -1/2(x-2)²  ↔ (y-k) = a(x-h)²

This parabola open downward (a = -1/2 <0), with a maximum shown in vertex
The vertex is (h , k) → Vertex(2 , 8)
focus(h, k +c )
 a = 1/4c → -1/2 = 1/4c → c = -1/2, hence focus(2, 8-1/2) →focus(2,15/2)

Latus rectum: y-value = 15/2

Replace in the equation y with 15/2→→15/2 = -1/2(x-2)² + 8
Or -1/2(x-2)² +8 -15/2 = 0
Solving this quadratic equation gives x' = 3 and x" = 2, then 
Latus rectum = 5

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The statement is wrong. a-b can be positive but not always.

Step-by-step explanation:

There can be three possible situations that are:

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Now taking first condition, the result will be positive because a smaller integer is subtracted from a greater one. Such as 5-2=3

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Solving systems of linear inequalities
PIT_PIT [208]

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YES. (2, 7) is a solution of the system.

Step-by-step explanation:

System of linear inequalities has been given as,

y ≥ -x + 1 --------(1)

y < 4x + 2 ------(2)

If (2, 7) is a solution of the given system of inequalities, it will satisfy both the inequalities.

By substituting the coordinates of point (2, 7) in inequality (1),

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7 < 4(2) + 1

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Therefore, point (2, 7) lie in the solution area of system of inequalities.

YES. (2, 7) is a solution of the system.

4 0
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An equilateral triangle with a side length of 6mm and a height of 5.2mm is divided vertically into halves. Find the area of one
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There are two ways you could go about solving this.

You could divide the length of the base (6mm) by 2 and use that to find the area or you could find the area of the whole triangle using 6mm and divide that by 2.

I will use the first method I described:

base = 6/2

base = 3 mm

height = 5.2 mm

area = bh/2

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(don't forget your units)


Using the other method would look like this:

area = bh/2

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area = (6 * 5.2)/2

area = 15.6 square mm

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As you can see either method yields the same result.

Hope this helped.

Cheers and good luck,

Brian

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