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ArbitrLikvidat [17]
3 years ago
15

I need help on this to make sure im doing it right.(3n)(6n^2) what is the answer

Mathematics
1 answer:
san4es73 [151]3 years ago
5 0
I believe this is the answer 18n^3
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Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
2 years ago
If 2 years on earth equals 5 months in another dimension, what are the equations to find the time in earth and that other dimens
Ede4ka [16]

Let x be the length of a month on the other dimension. Since a year value is 365.242 days

We have 2 × 365.242 = 5 × x

It means that 5x = 730,484

and that x = 730,484 / 5 ≈ 146,1 days rounded to the decimal

This means that a month length in the other dimension is of roughly 146 days.. and If we assume that a month in the other dimension is of 30 days approximately, then one day length would be of 146,1 / 30 ≈ 4.87 days

6 0
2 years ago
Which ordered pair is a solution of the equation y = 5x? (–2, 10) (–5, 25) (–3, 15) (–2, –10)
never [62]
The answer is (-2,-10) you just need to plug them in.
8 0
3 years ago
How would I solve for this ?
Daniel [21]
Multiply the length and width for a rectangle like the one shown.
5.3(2.6) = 13.78 square yards.
3 0
3 years ago
What is the 5004300 in standard form
Svet_ta [14]

Answer:

In standard form,

5004300. Five Millon four thousand and three hundred.

6 0
2 years ago
Read 2 more answers
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