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3 years ago

Solving by elimination -6x+6y=6 -6x+3y=-12

1 answer:

8 0

So elimination method is basically adding the equations and canceling out variables.
-6x + 6y = 6
-6x + 3y = -12
The eaiest way to solve is by multiplying the bottom equation by -1.
-6x + 6y = 6
6x - 3y = 12
Now you add the eqautions.
3y = 18
Divde 3 from both sides.
y = 6
Now plug in 6 into any of the original two equations. Lets use the first one.
-6x + 6(6) = 6
-6x + 36 = 6
Subtract 36 from both sides.
-6x = -30
Divide -6 from both sides.
x = 5
So your solution is (5, 6).
I hope this helps love! :)

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Which of the following equations have complex roots?

mestny [16]

Answer:

Step-by-step explanation:

Complex roots of quadratic functions occur when the <u>discriminant is negative</u>.

<u>Discriminant</u>

$b^2-4ac\quad\textsf{when}\:\:ax^2+bx+c=0$

Evaluate the discriminant of each of the given equations.

$\textsf{A.} \quad 3x^2+2=0$

$\implies a=3, \quad b=0, \quad c=2$

$\implies b^2-4ac=0^2-4(3)(2)=-24$

As -24 < 0 the equation will have complex roots.

$\textsf{B.} \quad 2x^2+1=7x$

$\implies 2x^2-7x+1=0$

$\implies a=2, \quad b=-7, \quad c=1$

$\implies b^2-4ac= (-7)^2-4(2)(1)=41$

As 41 > 0 the equation does not have complex roots.

$\textsf{C.} \quad 3x^2-1=6x$

$\implies 3x^2-6x-1=0$

$\implies a=3, \quad b=-6, \quad c=-1$

$\implies b^2-4ac=(-6)^2-4(3)(-1)=48$

As 48 > 0 the equation does not have complex roots.

$\textsf{D.} \quad 2x^2-1=5x$

$\implies 2x^2-5x-1=0$

$\implies a=2, \quad b=-5, \quad c=-1$

$\implies b^2-4ac=(-5)^2-4(2)(-1)=33$

As 33 > 0 the equation does not have complex roots.

Learn more about discriminants here:

brainly.com/question/27444516

brainly.com/question/27869538

Learn more about complex roots here:

brainly.com/question/26344541

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2 years ago

John used AABC to write a proof of the Centroid Theorem. He began by drawing medians AK and CL,

kumpel [21]

Answer:

Hello your question is poorly written attached below is the complete question

answer : attached below

Step-by-step explanation:

To Prove: Z is located 2/3 of the distance from each vertex of ΔABC to the midpoint of the opposite side. we will apply ; property of bisecting a line , equality theorem , transitive property and similarity theorem

Attached below is the proof