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goldfiish [28.3K]

2 years ago

13

You are building a toothpick fort for class. Each wall must be 1 1⁄3 feet long. Each toothpick is 2 inches long. How many toothp

icks will it take to make the base length of one wall

2 answers:

IceJOKER [234]2 years ago

8 0

Answer:

8 toothpicks

Step-by-step explanation:

Given: Each wall must be $1\frac{1}{3}$ feet long. Each toothpick is 2 inches long.

To find: How many toothpicks will it take to make the base length of one wall.

Solution:

We have, length of toothpick as 2 inches.

Length of wall $=1\frac{1}{3}$ feet.

Now, first we need to convert the length of wall in inches.

We know that $1\: \text{feet}=12\:\: \text{inches}$

So, $1\frac{1}{3} =\frac{4}{3}$ feet

$\frac{4}{3}=\frac{4}{3}\times12=16$ inches

Number of toothpicks will it take to make the base length of one wall $=\frac{\text{total length of wall}}{\text{length of each toothpick}}$

$=\frac{16}{2}$

$=8$

Hence, 8 toothpicks will be needed to make the base length of one wall.

Valentin [98]2 years ago

3 0

1 1/3 feet divided by 2 inches = 8

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Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

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Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

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The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

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We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

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