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s2008m [1.1K]
3 years ago
13

BIG POINTS Please solve. Thanks!

Mathematics
1 answer:
Nitella [24]3 years ago
3 0

9^1/3 * 3^x = 27^4/5

Rewrite 9 as 3^2

(3^2)^1/3 * 3^x = 27^4/5

Multiply the exponents in the first term:

3^2/3 * 3^x = 27^4/5

Use power rule to combine exponents:

3^(2/3 +x) = 27^4/5

Rewrite the 2nd term:

3^(2/3 +x) = (3^3)^4/5

Set the exponents only to equal:

2/3 + x = 3(4/5)

Solve for x:

simplify the right side:

2/3 + x = 12/5

Subtract 2/3 from both sides:

x = 26/15

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Not sure if you mean to ask for the first order partial derivatives, one wrt x and the other wrt y, or the second order partial derivative, first wrt x then wrt y. I'll assume the former.

\dfrac\partial{\partial x}(2x+3y)^{10}=10(2x+3y)^9\times2=20(2x+3y)^9

\dfrac\partial{\partial y}(2x+3y)^{10}=10(2x+3y)^9\times3=30(2x+3y)^9

Or, if you actually did want the second order derivative,

\dfrac{\partial^2}{\partial y\partial x}(2x+3y)^{10}=\dfrac\partial{\partial y}\left[20(2x+3y)^9\right]=180(2x+3y)^8\times3=540(2x+3y)^8

and in case you meant the other way around, no need to compute that, as z_{xy}=z_{yx} by Schwarz' theorem (the partial derivatives are guaranteed to be continuous because z is a polynomial).
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The measures of the angles of a convex polygon form an arithmetic sequence. The least measurement in the sequence is 128º. The g
Over [174]
\bf \qquad \qquad \textit{sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n}{2}(a_1+a_n)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\ a_n=n^{th}~value\\
----------\\
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\end{cases}

\bf S_n=\cfrac{n}{2}(128~+~172)\implies S_n=\cfrac{n}{2}(300)\implies S_n=150n
\\\\\\
\textit{but we also know that sum of all interior angles is }180(n-2)
\\\\\\
\stackrel{\textit{sum of the angles}}{S_n}=\stackrel{\textit{sum of the angles}}{180(n-2)}=150n
\\\\\\
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