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2 years ago

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Celine's Cereal Company is launching a new brand of cereal and she is considering two different sizes for the base of the boxes.

Box 1: The length is 3 times the width. Box 2: The length is 1 less than 4 times the width. The dimensions of the base of Box 1 are: width: x; length: 3 width: x; length: x + 3 width: x; length: x – 3 width: x; length: 3x The dimensions of the base of Box 2 are: width: x; length: 3 width: x; length: 4x – 1 width: x; length: x – 4 width: x; length: 4x + 1,
I have the answer for Box 1: width:x; legnth: 3x
but what is Box 3

2 answers:

Ronch [10]2 years ago

5 0

Answer

width: x; length: 4x – 1

Misha Larkins [42]2 years ago

5 0

Answer:

B. width: x; length: 4x – 1

Step-by-step explanation:

got it right on the assignment

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The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as

lubasha [3.4K]

Answer:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

$V = l * b * h$

Make h the subject of the formula

$h = \frac{V}{lb}$

The surface area (S) of the aquarium is:

$S = lb + 2(lh + bh)$

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

$C = 7 * lb + 1 * 2(lh + bh)$

$C = 7lb + 2(lh + bh)$

$C = 7lb + 2h(l + b)$

Substitute $\frac{V}{lb}$ for h in the above equation

$C = 7lb + 2*\frac{V}{lb}(l + b)$

$C = 7lb + \frac{2V}{lb}(l + b)$

$C = 7lb + \frac{2V}{b} + \frac{2V}{l}$

Differentiate with respect to l and with respect to b

$C_l=7b - \frac{2V}{l^2}$ $=0$

$C_b=7l - \frac{2V}{b^2}$ $=0$

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

$7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}$

$7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}$

By comparison:

$l =b$

$C_l=7b - \frac{2V}{l^2}$ $=0$ becomes

$7l - \frac{2V}{l^2}=0$

$7l = \frac{2V}{l^2}$

Cross Multiply

$7l^3 = 2V$

Solve for l

$l^3 = \frac{2V}{7}$

$l = \sqrt[3]{\frac{2V}{7}}$

Recall that: $l =b$

$b = \sqrt[3]{\frac{2V}{7}}$

Also recall that:

$h = \frac{V}{lb}$

$h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}$

$h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}$

Apply law of indices

$h = \sqrt[3]{\frac{49V^3}{4V^2}}$

$h = \sqrt[3]{\frac{49V}{4}}$

The dimension that minimizes the cost of material of the aquarium is:

$l = \sqrt[3]{\frac{2V}{7}}$ $b = \sqrt[3]{\frac{2V}{7}}$ $h = \sqrt[3]{\frac{49V}{4}}$

8 0

2 years ago

What dose 0.07x1.22 eaqual

iogann1982 [59]

0.07 * 1.22

Is equal to 0.0854

4 0

2 years ago

Read 2 more answers

Stacey has a square piece of cloth. She cuts 3 inches off of the length of the square and 3 inches off of the width. The area of

ratelena [41]

Answer:

The side length of the original square was 6 inches

Step-by-step explanation:

we know that

The area of a square is

$A=b^2$

where

b is the length side of the square

Let

x ---> the length of the original square

The area of the original square is

$A=x^{2}\ in^2$

The length of the smaller square is

$b=(x-3)\ in$

The area of the smaller square is

$A=(x-3)^2\ in^2$

The area of the smaller square is 1/4 the area of the original square

so

$(x-3)^2=\frac{1}{4} x^{2}$

solve for x

$x^2-6x+9=\frac{1}{4} x^{2}$

Multiply by 4 both sides

$4x^2-24x+36=x^{2}$

$4x^2-x^2-24x+36=0\\3x^2-24x+36=0$

Solve the quadratic equation by graphing

using a graphing tool

x=2, x=6

see the attached figure

The solution is x=6 in

Remember that the solution must be greater than 3 inches (because Stacey cuts 3 inches off of the length of the square and 3 inches off of the width)

therefore

The side length of the original square was 6 inches

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2 years ago

Deon plans to ride a 15-mi bicycle trail. If his average speed is 20 mi/h, which equation can he use to find the time t, in hour

fredd [130]

Answer:

The equation that Deon can use to find the time 't' it takes for his ride is t=distance/rate or t=15/20 or 0.75 hours.

Step-by-step explanation:

The relationship between distance, rate and time is a common formula represented by d=rt where 'd'=distance, 'r'=rate and 't'=time. Distance is typically calculated in miles, where rate is given in miles/hour and time is the hours. In this case, because we are looking for 't', we would simply plug in the values of distance and rate to get 20=15t. In order to solve for 't', we need to use inverse (opposite) operations, dividing both sides by 15 and getting 0.75 hours for 't'.

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2 years ago

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Lin and Jada each walk at a steady rate from school to the library. Lin can walk 13 miles in 5 hours, and Jada can walk 25 miles

astra-53 [7]

Answer:

It will take Lin an hour and 25 minutes at her constant speed, and it will take Jada at her constant speed an hour and 30 minutes.

SO, Lin will arrive at 4:25

and Jada will arrive at 4:30

There is a 5 minute difference in time.

Step-by-step explanation:

From the facts given, Lin can walk 13 miles in 5 hours which when you divide means she can walk 2.6 mph (miles per hour), and Jada can walk 2.5 mph.

To get these answers you divide the current speeds into the total distance

For example, in Lin's case...

3.25 (The 3 and 1/4 mile converted to decimal form)

2.6 (Lin's Average speed per hour)

3.25/2.6=1.25

1.25 (1 hour and 25 minutes)

5 0

2 years ago