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victus00 [196]
3 years ago
15

Find the area of the square rug if it is side is (7X +4) feet.

Mathematics
1 answer:
likoan [24]3 years ago
5 0

Answer:

49x^{2}+56x+16 ft squared.

Step-by-step explanation:

Alright, so we are given that we have a rug that has a side of 7x+4 feet. With this information, we are able to determine the area of the square rug. Considering that it is a square rug, all the sides are equal to each other because a square has four equal sides and angles.

To solve this, you can do it in two ways.

One, write it as (7x+4)2. That 2 is supposed to represent squared.

If you know the rule, this is a perfect square trinomial.

The formula is a squared plus 2ab plus b squared. 7 and 4 are your a and b values. 7x squared is 49x squared and 4 squared is 16.

2ab, multiply 2 times 7x times 4. That is 56x.

Therefore, the area is 49xsquared plus 56x+16.

Another way is to write it like this (7x+4) (7x+4).

Since it is an exponent, we are multiplying the same thing two times. Apply the Distributive Property.

E.G. 2(x+5)= 2x+10.

With this, you are distributing two numbers into another two numbers. 7x times 7x is 49x squared and 7x times 4 is 28x. Then, go to 4. 4 times 7x is also 28 x and 4 times 4 is 16. Now, combine the like terms. 49x squared plus 28x plus 28x or 56x plus 16.

This represents your area.

I hope this helps, and I hope you have a good day!

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If angle A = 34°, angle B = (x-5)°, angle C = 4y °, what is the value of y?
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180-34=146
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4 0
3 years ago
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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.

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