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vagabundo [1.1K]
3 years ago
6

Leia cuts congruent triangular patches with an area of 63 square centimeters from a rectangular piece of fabric that is 27 centi

meters long and 21 centimeters wide. How many of the patches can Leia cut from 33 pieces of the fabric?
Mathematics
1 answer:
m_a_m_a [10]3 years ago
6 0
<span>- The triangular patches have an area of 63 cm</span>² (A1= 63 cm²)<span>.
- The rectangular piece of fabric is 27 centimeters long and 21 centimeters wide. 

 1. Then, the first thing you must do is to calculate the area of the piece of fabric. The formula for the area of a rectangle is:

 A2=LxW
 
 L=27 centimeters (The lenght of the rectangle)
 W=21 centimeters (The width of the rectangle)

 2. When you substitute the values of L and W into the formula, you obtain:

 A2=(27 cm)(21 cm)
 A2=567 cm</span>²
<span>
 3. Then, let's calculate how many triangular patches she can cut from a rectangular piece of fabric:

 A2/A1= 567 cm</span>²/63 cm²= 9 triangular patches

 4. Finally, you can calculate how many triangular patches she can cut from 33 rectangular pieces of the fabric:

 (9 triangular patches)(33 pieces of fabric)=297 triangular patches
<span>
 How many of the patches can Leia cut from 33 pieces of the fabric?

 The answer is: 297 triangular patches</span>
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Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

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Step-by-step explanation:

Let's look at choice 1.

x=t+1

y=t^2+2t

I'm going to subtract 1 on both sides for the first equation giving me x-1=t. I will replace the t in the second equation with this substitution from equation 1.

y=(x-1)^2+2(x-1)

Expand using the distributive property and the identity (u+v)^2=u^2+2uv+v^2:

y=(x^2-2x+1)+(2x-2)

y=x^2+(-2x+2x)+(1-2)

y=x^2+0+-1

y=x^2

So this not the desired result.

Let's look at choice 2.

x=2t

y=4t^2+4t-3

Solve the first equation for t by dividing both sides by 2:

t=\frac{x}{2}.

Let's plug this into equation 2:

y=4(\frac{x}{2})^2+4(\frac{x}{2})-3

y=4(\frac{x^2}{4})+2x-3

y=x^2+2x-3

This is the desired result.

Choice 3:

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y=t^2+2t

Solve the first equation for t by adding 3 on both sides:

x+3=t.

Plug into second equation:

y=(x+3)^2+2(x+3)

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

y=(x^2+6x+9)+(2x+6)

y=(x^2)+(6x+2x)+(9+6)

y=x^2+8x+15

Not the desired result.

Choice 4:

x=t^2

y=2t-3

I'm going to solve the bottom equation for t since I don't want to deal with square roots.

Add 3 on both sides:

y+3=2t

Divide both sides by 2:

\frac{y+3}{2}=t

Plug into equation 1:

x=(\frac{y+3}{2})^2

This is not the desired result because the y variable will be squared now instead of the x variable.

Choice 5:

x=t+1

y=t^2+4t

Solve the first equation for t by subtracting 1 on both sides:

x-1=t.

Plug into equation 2:

y=(x-1)^2+4(x-1)

Distribute and use the binomial square identity used earlier:

y=(x^2-2x+1)+(4x-4)

y=(x^2)+(-2x+4x)+(1-4)

y=x^2+2x+-3

y=x^2+2x-3.

This is the desired result.

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Hey there! I'm happy to help!

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