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GarryVolchara [31]
3 years ago
10

What is the slope for -4x-3y=6

Mathematics
1 answer:
loris [4]3 years ago
7 0

Answer: y=4/3x+2


Step-by-step explanation:

-4x-3y=6

+4x       +4x

-3y=4x+6

-3/-3=4/3+6/3

y=4/3x+2


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Find the solution to the initial value problem that is a non-zero polynomial function in x. <img src="https://tex.z-dn.net/?f=xy
ohaa [14]
Bernoulli type.

xy'-15y=(4x^2-3x+7_y^{4/5}
xy^{-4/5}y'-15y^{1/5}=4x^2-3x+7

Let z=y^{1/5} so that z'=\dfrac15y^{-4/5}. Then the ODE becomes linear in z with

5xz'-15z=4x^2-3x+7
z'-\dfrac3xz=\dfrac45x-\dfrac35+\dfrac7{5x}
\dfrac1{x^3}z'-\dfrac3{x^4}z=\frac4{5x^2}-\dfrac3{5x^3}+\dfrac7{5x^4}
\left(\dfrac1{x^3}z\right)'=\frac4{5x^2}-\dfrac3{5x^3}+\dfrac7{5x^4}
\dfrac1{x^3}z=-\dfrac4{5x}+\dfrac3{10x^2}-\dfrac7{15x^3}+C
z=Cx^3-\dfrac45x^2+\dfrac3{10}x-\dfrac7{15}

y^{1/5}=Cx^3-\dfrac45x^2+\dfrac3{10}x-\dfrac7{15}
y=\left(Cx^3-\dfrac45x^2+\dfrac3{10}x-\dfrac7{15}\right)^5

Given that y(1)=0, we have

0=\left(C-\dfrac45+\dfrac3{10}-\dfrac7{15}\right)^5
\implies C=\dfrac{29}{30}

and so the particular solution is

y=\left(\dfrac{29}{30}x^3-\dfrac45x^2+\dfrac3{10}x-\dfrac7{15}\right)^5

Feel free to expand the solution to get it in the standard polynomial form.
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Answer:

image

Step-by-step explanation:

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3 years ago
What is the simplest form of ^4 sqrt 81x^8y^5
lara31 [8.8K]

ANSWER

3 {x}^{2} y \sqrt[4]{y}

EXPLANATION

We want to simplify:

\sqrt[4]{81 {x}^{8} {y}^{5}  }

We can split the radical sign to obtain:

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{5} }

Or

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}  \times y}

\sqrt[4]{81}  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}}  \times  \sqrt[4]{y}

\sqrt[4]{ {3}^{4} }  \times  \sqrt[4]{ {x}^{8} }  \times  \sqrt[4]{ {y}^{4}}  \times  \sqrt[4]{y}

Recall that:

\sqrt[n]{ {a}^{m} }  =  {a}^{ \frac{m}{n} }

{3}^{4 \times  \frac{1}{4} }  \times  {x}^{8 \times  \frac{1}{4}  }  \times  {y}^{4 \times  \frac{1}{4} }\times  \sqrt[4]{y}

3 {x}^{2} y \sqrt[4]{y}

6 0
4 years ago
Read 2 more answers
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