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Masteriza [31]
3 years ago
15

Jan is painting a square canvas. The length of one side of the canvas is 4 feet. Write an equation to show how to show how to fi

nd the area of the canvas
Mathematics
1 answer:
vlabodo [156]3 years ago
6 0

Step-by-step explanation:

It is given that, Jan is painting a square canvas. The length of one side of the canvas is 4 feet.

We need to write an equation to show how to find the area of the canvas.

For a square, all its sides are equal. Let the side is x. The area of a square is given by :

\text{Area}=\text{side}^2\\\\A=x^2\\\\A=4^2\\\\A=16

So, the equation that shows the area of the canvas is A=x^2.

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1/6

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Any line with the same slope. In this case the slope is 1. For example, y=x+4 is parallel to y=x+7.

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I NEED HELP PLZ I WILL GIVE THANKS AND BRAINLIST
ratelena [41]

Answer:

Point B

Step-by-step explanation:

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3 years ago
Solve the equation
NNADVOKAT [17]

Answer:

<em>x = 0.04979</em>

Step-by-step explanation:

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8 0
2 years ago
Read 2 more answers
Help!! 50 points and brainliest!
Viktor [21]

Answer:

Second choice:

x=2t

y=4t^2+4t-3

Fifth choice:

x=t+1

y=t^2+4t

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:

x=t-3

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.

3 0
3 years ago
Read 2 more answers
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