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chubhunter [2.5K]

1 year ago

12

An elevator starts at floor 5. It goes up 6 floors twice, and then it goes down 8 floors. A second elevator

2 answers:

omeli [17]1 year ago

6 0

Answer:

Part 1 is ELevator 2 because elevator one ends on floor 9 and elevator 2 ends on floor 23. Part 2 is Elevator 1 ends on floor 9 and Elevator 2 ends on floor 23.

Step-by-step explanation:

astraxan [27]1 year ago

6 0

Step-by-step explanation:

first elevator is on floor 9 other one is on floor 23 i belive so the 2nd one is higher

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Graph the line with slope 3 passing through the point (5,-4).

12345 [234]

Answer:

y = 3x-16

Step-by-step explanation:

two points where the line goes through is (5,-4) as mentioned and (0,-16)

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3 years ago

The equation 2x2 − 12x + 1 = 0 is being rewritten in vertex form. Fill in the missing step. Given 2x2 − 12x + 1 = 0 Step 1 2(x2

Bezzdna [24]

Answer:

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

Part 5) x = 6, x = 44; The zeros represent the number of monthly memberships where no profit is made

Part 6) 2(x − 7)2 + 118; x = $7

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

Part 8) (x + 3)^2 − 4

Part 9) 2(x − 1)^2 = 4

Part 10) 8(x − 4)^2 + 592

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

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

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

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

Rewrite as perfect squares

$(x-25)^{2}=361$

square root both sides

$(x-25)=(+/-)19$

$x=25(+/-)19$

$x=25(+)19=44$

$x=25(-)19=6$

Part 6) we have

$-2x^{2}+28x+20$

This is a vertical parabola open downward

The vertex is a maximum

Convert the equation into vertex form

Factor the leading coefficient

$-2(x^{2}-14x)+20$

Complete the square

$-2(x^{2}-14x+49)+20+98$

$-2(x^{2}-14x+49)+118$

Rewrite as perfect square

$-2(x-7)^{2}+118$

The vertex is the point (7,118)

therefore

The video game price that produces the highest weekly profit is x=$7

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

The maximum height of the puck is 4 feet.

Part 8) we have

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

Convert to vertex form

Group terms

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

Complete the square

$(x^{2}+6x+9)+5-9$

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

Rewrite as perfect squares

$(x+3)^{2}-4$

Part 9) we have

$2x^{2}-4x-2=0$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 2 the leading coefficient

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

Complete the square

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

$2(x^{2}-2x+1)-4=0$

Rewrite as perfect squares

$2(x-1)^{2}-4=0$

$2(x-1)^{2}=4$

The vertex is the point (1,-4)

Part 10) we have

$8x^{2}-64x+720$

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert to vertex form

Factor 8 the leading coefficient

$8(x^{2}-8x)+720$

Complete the square

$8(x^{2}-8x+16)+720-128$

$8(x^{2}-8x+16)+592$

Rewrite as perfect squares

$8(x-4)^{2}+592$

the vertex is the point (4,592)

The population has a minimum at x=4 years ( that is after 4 years since 1998 )

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2 years ago

John is hiking a trail that is 0.8 miles long he hikes the first 2/10 of the distance by himself he hates the rest of the way wi

jenyasd209 [6]

I think the answer is .16

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3 years ago

Find f(3) for f(x) = 1/2(4)^x <br> A. 8<br> B. 32<br> C. 6<br> D. 64

Elza [17]

1/2 (4)^3

1/2 x 64

32

7 0

3 years ago

Read 2 more answers

Helppppp with this. What do I have to do?

vaieri [72.5K]

You need to use the grids to draw rectangles which meet the requirements to help you put the answers in on the right.

1) Short side of 7 and long side of 9.

2) Short side of 9 and long side of 10.

3) Short side of 3 and long side of 8.

4) Short side of 4 and long side of 9.

5) Short side of 3 and long side of 5.

6) Short side of 6 and long side of 8.

7) Short side of 1 and long side of 2

8) Both sides 10 (this is technically a square).

9) Short side of 6 and long side of 9.

10) Short side of 7 and long side of 8.

11) Short side of 2 and long side of 3.

12) Short side of 6 and long side of 9.

You can draw these a few different ways to still get a correct result, so above are just one way of doing it.

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3 years ago