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3 years ago

A separable differential equation is a first-order differential equation that can be algebraically manipulated to look like:

1 answer:

iogann1982 [59]3 years ago

4 0

<span>A separable differential equation is a first-order differential equation that can be algebraically manipulated to look like:

A) f(y)dy = g(x)dx

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>

You might be interested in

Explain why any of the four operations placed between two terms 5 and -3 sqrt (8) will result in an irrational number

Gnoma [55]

Sum/difference:

Let

$x = 5 + (-3\sqrt{8}) = 5-3\sqrt{8}$

This means that

$3\sqrt{8} = 5-x \iff \sqrt{8} = \dfrac{5-x}{3}$

Now, assume that $x$ is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that $x$ was rational, which proves that the sum/difference of the two given terms was irrational

Multiplication/division:

The logic is actually the same: if we multiply the two terms we get

$x = -15\sqrt{8}$

if again we assume x to be rational, we have

$\sqrt{8} = -\dfrac{x}{15}$

But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.

7 0

2 years ago

What is the value of K?

Ksju [112]

Answer: The value of k for which one root of the quadratic equation kx2 - 14x + 8 = 0 is six times the other is k = 3.
Let's look into the solution step by step.

Explanation:

Given: A quadratic equation, kx2 - 14x + 8 = 0

Let the two zeros of the equation be α and β.

According to the given question, if one of the roots is α the other root will be 6α.

Thus, β = 6α

Hence, the two zeros are α and 6α.

We know that for a given quadratic equation ax2 + bx + c = 0

The sum of the zeros is expressed as,

α + β = - b / a

The product of the zeros is expressed as,

αβ = c / a

For the given quadratic equation kx2 - 14x + 8 = 0,

a = k, b = -14, c = 8

The sum of the zeros is:

α + 6α = 14 / k [Since the two zeros are α and 6α]

⇒ 7α = 14 / k

⇒ α = 2 / k --------------- (1)

The product of the zeros is:

⇒ α × 6α = 8 / k [Since the two zeros are α and 6α]

⇒ 6α 2 = 8 / k

⇒ 6 (2 / k)2 = 8 / k [From (1)]

⇒ 6 × (4 / k) = 8

⇒ k = 24 / 8

⇒ k = 3

6 0

2 years ago

Solve the equation<br> x-5=36-7(x+7)

lana66690 [7]

Answer: x = -1

Step-by-step explanation:

1) x - 5 = 36 -7x -49

2) x - 5 = -13 -7x

3) x + 7x =- 13 +5

4) 8x =- 8

5) x = -1

Please mark this as the brainliest answer! :) I hope this helped! <3

6 0

3 years ago

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}$

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

3 years ago

Solve 9(5-21/7)+4(2)

masya89 [10]

Answer:

Hiiiiiiii

The answer is 26

5 0

2 years ago