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3 years ago

3/4x-4=5+x Helppppp meeeeee

Mathematics

1 answer:

Viefleur [7K]3 years ago

4 0

Answer:

x = -36

Step-by-step explanation:

3/4x-4=5+x

Subtract 3/4x from each side

3/4x - 3/4x-4=5+x -3/4x

3/4x - 3/4x-4=5+4/4x -3/4x

-4 = 5 + 1/4x

Subtract 5 from each side

-4-5 = 1/4x

-9 = 1/4x

Multiply each side by 4

4 * -9 =1/4x*4

-36 =x

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3 years ago

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3 years ago

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3 years ago

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3 years ago

Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$