Solve this problem: 125^-1/3×49^-1/2

A population of rabbits on a farm grows by 12% each year. Assume that this increase occurs as new baby rabbits at the beginning

jolli1 [7]

Answer:

a) r₁₂ = 104.36

In general, rₙ = arⁿ⁻¹

b)

- rabbit food consumed during the 10th year is approximately 832 pounds

- rabbit food consumed in total for the 1st through 10th years is approximately 5265 pounds

Step-by-step explanation:

Given that:

r1 = 30 and a farm grows by 12%

a = 30 and the common ratio r = 1.12

now

n r

1 30.00

2 33.60

3 37.63

4 42.15

5 47.21

6 52.87

7 59.21

8 66.32

9 74.28

10 83.19

11 93.18

12 104.36

Therefore r₁₂ = 104.36

In general, rₙ = arⁿ⁻¹

b)

if each rabbit consume 10 lbs of rabbit food each year

n r food consumed(lbs)

1 30.00 300

2 33.60 336

3 37.63 376

4 42.15 422

5 47.21 472

6 52.87 529

7 59.21 592

8 66.32 663

9 74.28 743

10 83.19 832

total 5265

Therefore, the rabbit food consumed during the 10th year is approximately 832 pounds

And the rabbit food consumed in total for the 1st through 10th years is approximately 5265 pounds

6 0

2 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 )

6 0

3 years ago

Find the slope and initial value of the graph below. Then write the equation of the line. Write the equation of the line.

LenaWriter [7]

Answer:

y= x + 2

Hope this helps.

7 0

3 years ago

12 grapes cost 90 $how many can bought in 40$

likoan [24]

Let x = the amount of grapes that can be bought for 0.40.

Note: By 90 I assume you mean 90 cents. The same for 40.

12/x = 0.90/0.40

0.90x = 12(0.40)

0.90x = 4.80

x = 4.80 ÷ 0.90

x = 5.33333

We can round the decimal to the ones place to get 5 grapes.

4 0

3 years ago

Read 2 more answers

I forgot how to solve this question <br> 7x +9=58

daser333 [38]

Minus 9 from both sides

7x = 49

X = 7

6 0

3 years ago

Read 2 more answers

Solve this problem: 125^-1/3×49^-1/2 ​

Solve this problem: 125^-1/3×49^-1/2