hi! i dont have the answer because i suck at school lol XD but, i actually have the same question! do you know the answer yet?:] sorry to bother you if you dont.

Step-by-step explanation:

3 0

3 years ago

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Create an equivalent system of these equations and test your solution x + y = 1 x 3y =9

ad-work [718]

An equvilent equation
remember you can do anything to an equation as long asyou do it to both sides

assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent

ok, solve initial

x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation

-x-y=-1
x-3y=9 +
0x-4y=8

-4y=8
divide both sides by -4
y=-2

sub back
x+y=1
x-2=1
add 2
x=3

x=3
y=-2
(3,-2)

if we test it in other one

2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep

2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep

solution is (3,-2)

4 0

3 years ago

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Simplify the following radical Sqrt 126x^13

fenix001 [56]

Answer:

$\large\boxed{\sqrt{126x^{13}}=3x^6\sqrt{14x}}$

Step-by-step explanation:

$Domain:\ x\geq0\\\\\sqrt{126x^{13}}=\sqrt{9\cdot14\cdot x^{12+1}}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{9\cdot14\cdot x^{12}\cdot x^1}=\sqrt{9\cdot14\cdot x^{6\cdot2}\cdot x}\\\\\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\ \text{and}\ (a^n)^m=a^{nm}\\\\=\sqrt9\cdot\sqrt{14}\cdot\sqrt{(x^6)^2}\cdot\sqrt{x}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=3\cdot\sqrt{14}\cdot x^6\cdot\sqrt{x}=3x^6\sqrt{14x}$

3 0

3 years ago