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3 years ago

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

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

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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

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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