Ask question

Ask question

All categories

2 years ago

12

If ABC and DEF are supplementary, then ABC and DEF form a linear pair

1 answer:

-Dominant- [34]2 years ago

4 0

What are you asking? If you are asking if thats correct then that is only true if they are adjacent or right next to each other.

You might be interested in

Given the linear equation r=-2.5 t and t can only be a negative number, what quandrant will the r coordinate be located?

Citrus2011 [14]

Answer:

quadrant 4 i think

Step-by-step explanation:

4 0

1 year ago

Read 2 more answers

PLEASE HELP ME WITH THESE QUESTIONS!!

Varvara68 [4.7K]

Answer:

3. [1, −2]

2. [−3, 3]

1. [−7, 10]

Step-by-step explanation:

3.

{7⁄2x - ½y = 9⁄2

{3x - y = 5

-6⁄7[7⁄2x - ½y = 9⁄2]

{−3x + 3⁄7y = −3 6⁄7 >> New Equation

{3x - y = 5

_________________

-4⁄7y = 1 1⁄7

-2 = y [Plug this back into both equations above to get the x-coordinate of 1]; 1 = x

__________________________________________________________

2.

{−3x + 9y = 36

{4x + 12y = 24

¾[4x + 12y = 24]

{−3x + 9y = 36

{3x + 9y = 18

______________

18y = 54

___ ___

18 18

y = 3 [Plug this back into both equations above to get the x-coordinate of −3]; −3 = x

__________________________________________________________

1.

{4x − y = −38

{x + y = 3

_____________

5x = -35

___ ____

5 5

x = -7 [Plug this back into both equations above to get the y-coordinate of 10]; 10 = y

I am joyous to assist you anytime.

3 0

3 years ago

I dont get this please help i will mark the best

Mumz [18]

/Answer:

the verry last one

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Y=2(x+4)^2-21<br><br> Convert the equation from vertex form to standard

PSYCHO15rus [73]

Answer:

y = 2x² + 16x + 11

I hope im right... If not reaaally sorry!!

7 0

2 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

1 year ago