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3 years ago

7

i catch a lot of people taking other peoples answer here's one for instance https://brainly.com/question/14903527 my question is

at the bottom so you can see the difference Extra Points :)

1 answer:

Lisa [10]3 years ago

4 0

Answer:

he pep

Step-by-step explanation:

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Solve for substitution<br> x-2y+z=-11<br> 3x+2y-z=7<br> -x+2y+4z=-9

Alla [95]

Answer:

Unit rate miles per day is, 3

Step-by-step explanation:

Unit rate are defined as the rate are expressed as quantity of 1 such that 2 miles per hour or 5 km per hour.

As per the given statement: A construction crew built 1/2 miles of road in 1/6 days

by unit rate definition;

After substituting the given values we have

⇒

or

Therefore, unit rate miles per day is, 3

Step-by-step explanation:

hhhthhg

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3 years ago

Is the sum even or odd? 6+7

valentina_108 [34]

Odd.................................

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3 years ago

Requie<br> 2) 20 + (-45)

harkovskaia [24]

Answer:

20 + (-45) = -25

6 0

3 years ago

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A time series that shows a recurring pattern over one year or less is said to follow a Select one:______

zaharov [31]

The answer is C. Seasonal pattern

3 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