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erik [133]
3 years ago
6

Two boxes have the same volume. Box 1 has dimensions of 8 ft. by 4 ft. by 3 ft. Box 2 has dimensions of 12 ft. by 4ft by 2 ft. T

he material to make the boxes cost $3.00 per square foot. How much more does Box 2 cost than Box 1?
Mathematics
1 answer:
hram777 [196]3 years ago
3 0
You would have to find the surface areas of each box, which would be 8*4 *2, 8*3 *2, and 4*3 *2. That *3, would be the cost, and then surface area of 2 would be 12*4 *2, 12*2 *2, and 2*4 *2. That *3 would again be the cost to make the box. That all would be A=64 + 48 + 24, or 136*3 = $408 to make box A, and box B=96 + 48 + 16 = 160*3 = $480 to make box B. Therefore, box B, or $480 - $408 = $72 more.
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Elena bikes 20 minutes each day for exercise. Write an equation to describe the relationship between her distance in miles, D, a
kompoz [17]

<u>Answer:</u>

The equation to describe the relationship between Elena distance:

a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

<u>Solution:</u>

Given, Elena bikes 20 minutes each day for exercise.  

We have to write an equation to describe the relationship between her distance in miles, D, and her biking speed, in miles per hour,  

We know that, distance travelled = speed \times time

a. At a constant speed of 13 miles per hour for the entire 20 minutes  

Her speed is 13 miles per hour.  

Then, distance D miles = 13 miles per hour \times 20 minutes

\mathrm{D}=13 \text { miles per hour } \times \frac{20}{60} \text { hours } \rightarrow \mathrm{d}=13 \times \frac{1}{3} \rightarrow \mathrm{d}=4.34 \text { miles approximately. }

b. At a constant speed of 15 minutes per hour for the first 5 minutes, then at 12 miles per hour for the last 15 minutes  

Now, total distance travelled = distance travelled with 15 mph + distance travelled with 12 mph

\begin{array}{l}{\mathrm{D}=15 \mathrm{mph} \times 5 \text { minutes }+12 \mathrm{mph} \times 15 \text { minutes }} \\\\ {\mathrm{D}=15 \mathrm{mph} \times \frac{5}{60} \text { hours }+12 \mathrm{mph} \times \frac{15}{60} \text { minutes }} \\\\ {\mathrm{D}=15 \times \frac{1}{12}+12 \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{5}{4}+3 \rightarrow \mathrm{D}=3+1.25 \rightarrow \mathrm{D}=4.25 \mathrm{miles}}\end{array}

c. At a constant speed of M miles per hour for the first 5 minutes, then at N miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with M mph + distance travelled with N mph

D = M mph \times 5 minutes + N mph \times 15 minutes

\mathrm{D}=\mathrm{M} \mathrm{mph} \times \frac{5}{60} \text { hours }+\mathrm{N} \mathrm{mph} \times \frac{15}{60} \text { minutes }

\mathrm{D}=\mathrm{M} \times \frac{1}{12}+\mathrm{N} \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{N}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{3 N}{12} \rightarrow 12 \mathrm{D}=\mathrm{M}+3 \mathrm{N} \text { miles }

Hence, a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

8 0
4 years ago
Before the industrial revolution in 1800 the concentration of carbon in Earth’s atmo- sphere was 280 ppm. The concentration in 2
slega [8]

Answer: There is increase of 4.255 in the amount of carbon in the atmosphere.

Step-by-step explanation:

Since we have given that

Concentration of carbon in Earth's atmosphere in 1800 = 280 ppm

Concentration of carbon in Earth's atmosphere in 2015 = 399 ppm

We need to find the percentage increase in the amount of carbon in the atmosphere.

So, Difference = 399-280 = 119 ppm

so, percentage increase in the amount of carbon is given by

\dfrac{Difference}{Original}\times 100\\\\=\dfrac{119}{280}\times 100\\\\=\dfrac{11900}{280}\\\\=42.5\%

Hence, there is increase of 4.255 in the amount of carbon in the atmosphere.

4 0
3 years ago
A student is 5 feet, 2 inches tall. What is her height in meters?
Maurinko [17]
We know that 1 feet is 0.3m and that 1 inch = 0.0254 m
To get the result we have to multiply 5 feet by 0.3 and 2 inches by 0.0254 and add it.
5*0.3=1.5
2*0.0254=0.0508
Now adding
1.5+0.058=1.55m - its the answer
6 0
4 years ago
Jose practiced piano 5 days this week. He practiced 3/4 of an hour each day.
emmainna [20.7K]

Answer:

3 hours and 15 min is how long he practiced

3 0
3 years ago
Read 2 more answers
Simplifyng algebraic expressions<br><br>-2z-3w+7+8 use z=3 and w=9​
Anettt [7]

Answer:

-9

Step-by-step explanation:

Substitute 3 for <em>z</em><em> </em>and 9 for <em>w</em> in the equation.

- 2z - 3w + 7 + 8

- 2(3) - 3(9) + 15

- 6 - 18 + 15

- 24 + 15

- 9

7 0
3 years ago
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