You have to switch the [latex]x[/latex] and [latex]y[/latex], and then solve for [latex]y[/latex].
[latex]y=x^7+x^5 [/latex]
[latex]x=y^7+y^5 [/latex]
[latex]x=y^5(y^2+1) [/latex]
I think it can't be solved farther.
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Use order of operation, i.e.,PEMDAS (Parenthesis, Exponent, Multiplication, Division, Addition, Subtraction)
[latex] 36\div 6(2+2+2)= ? [/latex]
First, Parenthesis-
[latex] =36\div6(6) [/latex]
Now, between multiplication and division, we apply operation which comes first, here division comes first so we apply division-
[latex]= 6(6) [/latex]
Now, applying multiplication-
[latex]= 6(6)\;\;\implies 6\times6=36 [/latex]
I hope it' ll help.
We substitute [latex] \sqrt{9-x^2} [/latex] by [latex] x= 3sin\theta [/latex] because a trigonometric substitution</a> helps us to get a perfect square under the radical sign. This simplifies the integrand function.http://www.actucation.com/calculus-2/indefinite-integration/basic-methods-of-integration/integration-by-substitution-of-trigonometric-functions
Now, we can simplified it easily,
[latex] \int\sqrt{9-x^2}\;\;\implies\int\sqrt{3^2-x^2} [/latex]
[latex] \int\sqrt{3^2-x^2}\;\implies\sqrt{3^2-(3sin\theta)^2} [/latex]
[latex] =\sqrt{9-9sin^2\theta} [/latex]
[latex] \sqrt{9-9sin^2\theta}\;\;\implies\sqrt{9(1-sin^2\theta)}\;\;\implies\sqrt{9cos^2\theta}\;=\;\int 3cos\theta [/latex]
I hope it' ll help.
You can simplify it easily,
[latex] \frac{48}{2y} [/latex]
Write this as,
[latex] \frac{48}{2}\times\frac{1}{y} [/latex]
Now, simply divide 48 by 2 and then multiply,
[latex] 24\times\frac{1}{y}=\frac{24}{y} [/latex]
I hope it' ll help.
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