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3 years ago

7

PLEASE HELPPPP

1 answer:

valkas [14]3 years ago

3 0

Answer:

$y = -\frac{5}{2}x + 6$

Step-by-step explanation:

Using a point (4, -4) and the slope of the line that passes through it, m = -⁵/2, we can generate an equation by using the point-slope form equation of a line given as y - b = m(x - a).

Where,

(a, b) = (4, -4)

m = -⁵/2

Plug in the values

y - (-4) = -⁵/2(x - 4)

y + 4 = (-5x + 20)/2

Multiply both sides by 2

2y + 8 = -5x + 20

Subtract both sides by 8

2y = -5x + 20 - 8

2y = -5x + 12

Divide both sides by 2

y = -⁵/2x + 12/2

$y = -\frac{5}{2}x + 6$ (equation in slope-intercept form)

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Part 1) $2(x-3)^{2}-17=0$ (the missing steps in the explanation)

Part 3) (8, 4); The vertex represents the maximum profit

Part 4) x = 3.58, 0.42

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Part 6) 2(x − 7)2 + 118; x = $7

Part 7) The maximum height of the puck is 4 feet. −(x − 4)^2 + 6

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Step-by-step explanation:

Part 1) we have

$2x^{2} -12x+1=0$

Convert to vertex form

step 1

Factor the leading coefficient and complete the square

$2(x^{2} -6x)+1=0$

$2(x^{2} -6x+9)+1-18=0$

step 2

$2(x^{2} -6x+9)+1-18=0$

$2(x^{2} -6x+9)-17=0$

step 3

Rewrite as perfect squares

$2(x-3)^{2}-17=0$

Part 3) we have

$f(x)=-x^{2}+16x-60$

we know that

This is the equation of a vertical parabola open downward

The vertex is a maximum

Convert to vertex form

$f(x)+60=-x^{2}+16x$

Factor the leading coefficient

$f(x)+60=-(x^{2}-16x)$

Complete the squares

$f(x)+60-64=-(x^{2}-16x+64)$

$f(x)-4=-(x^{2}-16x+64)$

Rewrite as perfect squares

$f(x)-4=-(x-8)^{2}$

$f(x)=-(x-8)^{2}+4$

The vertex is the point (8,4)

The vertex represent the maximum profit

Part 4) Solve for x

we have

$-2(x-2)^{2}+5=0$

$-2(x-2)^{2}=-5$

$(x-2)^{2}=2.5$

square root both sides

$(x-2)=(+/-)1.58$

$x=2(+/-)1.58$

$x=2(+)1.58=3.58$

$x=2(-)1.58=0.42$

Part 5) we have

$f(x)=-x^{2}+50x-264$

we know that

The zeros or x-intercepts are the value of x when the value of the function is equal to zero

so

In this context the zeros represent the number of monthly memberships where no profit is made

To find the zeros equate the function to zero

$-x^{2}+50x-264=0$

$-x^{2}+50x=264$

Factor -1 of the leading coefficient

$-(x^{2}-50x)=264$

Complete the squares

$-(x^{2}-50x+625)=264-625$

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$(x^{2}-50x+625)=361$

Rewrite as perfect squares

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$x=25(+/-)19$

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Convert the equation into vertex form

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Rewrite as perfect square

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Part 7) we have

$f(x)=-x^{2}+8x-10$

Convert to vertex form

$f(x)+10=-x^{2}+8x$

Factor -1 the leading coefficient

$f(x)+10=-(x^{2}-8x)$

Complete the square

$f(x)+10-16=-(x^{2}-8x+16)$

$f(x)-6=-(x^{2}-8x+16)$

Rewrite as perfect square

$f(x)-6=-(x-4)^{2}$

$f(x)=-(x-4)^{2}+6$

The vertex is the point (4,6)

therefore

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Part 8) we have

$x^{2}+6x+5$

Convert to vertex form

Group terms

$(x^{2}+6x)+5$

Complete the square

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$(x^{2}+6x+9)-4$

Rewrite as perfect squares

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Convert to vertex form

Factor 2 the leading coefficient

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Complete the square

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Rewrite as perfect squares

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Factor 8 the leading coefficient

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Rewrite as perfect squares

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