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pickupchik [31]
2 years ago
7

Solve the equation for y. 6 = 4x + 9y

Mathematics
2 answers:
faltersainse [42]2 years ago
7 0

Answer:

-4/9x + 2/3 =Y.

I hope this

is the answer

k0ka [10]2 years ago
6 0
Subtract 4x from both sides
Then divide both sides by 9 to isolate y

Y=(6-4x)/9
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Citrus2011 [14]

Answer:

Solving quadratic equations can be difficult, but luckily there are several different methods that we can use depending on what type of quadratic that we are trying to solve. The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.

Step-by-step explanation:

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3 0
3 years ago
1. Below is data collected from a random sample of 40 people at the public library. If the public library has 300 people, then w
nataly862011 [7]

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please do the math luke a lesson and the beta is very weird so you want to do something about that so i suggest that you do 300.

5 0
3 years ago
Can anyone please help me on this?
Elena L [17]

Answer:

b

Step-by-step explanation:

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3 0
3 years ago
Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra
Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize x^2-\frac13 as yet another difference of squares:

x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)

And if you're working over the field of complex numbers, you can go even further. For instance,

x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)

But I think you'd be fine stopping at the first result,

x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}

6 0
3 years ago
Write the missing exponent of 100=10 =
Ganezh [65]
Maybe 5 is the answer
4 0
3 years ago
Read 2 more answers
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