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Oksanka [162]
3 years ago
5

4x+4 jbujgbujbjkgubujgbn jbguhvnhghjf

Mathematics
1 answer:
Advocard [28]3 years ago
3 0

Answer: You can’t really add them

Step-by-step explanation: Since 4x has a variable, and 4x and 4 aren’t like terms, you CAN NOT add them together. I’m not sure if that was what you were looking for.

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kevin opens a savings account with $750. the interest rate is 5%, compounded annually. How much money will he have in his accoun
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over the weekend and mom went out with her family they went to the morning Cafe while there they spent $44 her mom had a discoun
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price after discount- $44-$11=$33

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5 0
3 years ago
Solve the systems of equations y=2x and y=-x - 6
slava [35]

Answer:

y = -4

x = -2

Step-by-step explanation:

Hi there...

y = 2x

y = -x - 6

Since why is equal to 2x, we can plug it in for y in the second equation. Here's what I mean...

2x = -x - 6

add x on both sides...

3x = -6

divide by 3 on both sides...

x = -2

Now, plug in x for y. Here's what I mean...

y = 2x

y = 2 (-2)

y = -4

Hope this helps :)

7 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
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All u have to do is add 23.5 to 23.5
7 0
3 years ago
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