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Inga [223]
3 years ago
5

Why is (x^2+2xy+y^2) greater than 2(x+y) when x and y are over zero and X>Y

Mathematics
1 answer:
VikaD [51]3 years ago
6 0

Answer:

Step-by-step explanation:

This is true because if we were to plug in values for x and y the first equation will have a greater value.

Lets try 2 for x and y......

2^2+2(2)(2)+2^2                                                                                              

4+8+4

=16

and for the other expression

2(2+2)

4+4

8

And 16 is greater then 8 this is true for all real rational numbers

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Express ✓ – 75 in its simplest form.
timofeeve [1]

Step-by-step explanation:

The correct answer is 5i\sqrt{3}

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2 years ago
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mafiozo [28]
Hehe lovely what the
8 0
3 years ago
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
$750 are deposited into an account quarterly for eight yeara at an interest rate of 7.9% compounded quarterly how much is in the
IgorC [24]
750*7.9%*8=474 is the account at the end of 8 years. Good luck
4 0
4 years ago
Find the product. Write your answer in scientific notation. (3 x 10^2)×(4×10^5) ​
suter [353]

Answer:

12 * 10 ^ 7

Step-by-step explanation:

3 * 10 ^ 2 * 4 * 10 ^5

= 12 * 10^ 2 + 5

= 12 * 10 ^ 7

Hope it will help :)

7 0
3 years ago
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