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2 years ago

Find the equation that is perpendicular to 7y= -3x passing through (9,1).

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

8 0

Answer:y=7/3x−20. that is the equation that is perpendicular.

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Sindrei [870]

Answer:D

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3 years ago

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viktelen [127]

Answer:

G and E

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2 years ago

Write a subtraction problem involving regrouping that has ted reading 304 pages. Answer your question

kari74 [83]

How about this:

The book has 712 pages. Ted has already read 408 pages, so he has 304 left to read.
712 - 408 = 304

4 0

2 years ago

8•x + 10 +2x +4 •x + 8 -6

Tcecarenko [31]

Answer:

14x + 12

Step-by-step explanation:

8 times x would be 8x

4 times x would be 4x

8x+10+2x+4x+8-6

8x+2x+4x=14x

10+8-6=12

=14x + 12

5 0

2 years ago

Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.

Mama L [17]

Multiplying both sides by $y^2$ gives

$xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1$

so that substituting $v=y^3$ and hence $\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}$ gives the linear ODE,

$\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1$

Now multiply both sides by $3x^2$ to get

$x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2$

so that the left side condenses into the derivative of a product.

$\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2$

Integrate both sides, then solve for $v$ , then for $y$ :

$x^3v=\displaystyle\int3x^2\,\mathrm dx$

$x^3v=x^3+C$

$v=1+\dfrac C{x^3}$

$y^3=1+\dfrac C{x^3}$

$\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}$

6 0

3 years ago