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3 years ago

Please answer this correctly

Mathematics

2 answers:

LekaFEV [45]3 years ago

8 0

Answer:

Mean

Step-by-step explanation:

scoray [572]3 years ago

7 0

The correct answer to this is mean

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The answer is r and s because it’s in the same plane.

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3 years ago

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Braceleftzf y = 8/7 x − 2/3 <br> 8/3 x + y = 5/9

givi [52]

Answer:

(x, y) = (77/240, -3/10)

Step-by-step explanation:

It is convenient to write the equations in standard form.

Multiplying the first equation by 21 gives ...

21y = 24x -14

Multiplying the second equation by 8 gives ...

24x +9y = 5

Then the system of equations in standard form is ...

24x -21y = 14
24x +9y = 5

Subtracting the first from the second, we get ...

(24x +9y) -(24x -21y) = (5) -(14)

30y = -9

y = -9/30 = -3/10

Substituting this into the second equation, we have ...

24x +9(-3/10) = 5

24x = 7.7 . . . . . . . add 27/10

x = 7.7/24 = 77/240

The solution is (x, y) = (77/240, -3/10).

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3 years ago

PLEASE HELP WITH GRAPHING QUESTION ASAP!!

bulgar [2K]

x-int: -0.5; y-int: 1

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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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3 years ago

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