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3 years ago

5

Simplify this for me thanks u

1 answer:

almond37 [142]3 years ago

3 0

$4n + 3(n - 5) + 2 \\ \\ = 4n + 3n - 15 + 2\\ \\ = 7n - 13$

<h3>Option B</h3>

<h3>Hope This Helps You ❤️</h3>

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oee [108]

Answer:

yes it is parallel

Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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2 years ago

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Lf f(x) = 5x, what is f^-1(x)?

Nonamiya [84]

Answer: $f^{-1}(x)=\frac{x}{5}$

Step-by-step explanation:

By definition the domain of an inverse function $f^-1(x)$ is the range of f(x) and the range of the inverse function is equal to the domain of the principal function f(x).

If you have a function $f(x)=5x$ , then to find the inverse function, follow these steps:

1. Make $y=f(x)$

$f(x)=y=5x$

$y=5x$

2. Solve for the variable "x":

$x=\frac{y}{5}$

3. Exchange the variable "x" with the variable "y":

$y=\frac{x}{5}$

4. Exchange "y" with $f^{-1}(x)$ . Then the inverse function is:

$f^{-1}(x)=\frac{x}{5}$

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3 years ago

162*0.967=what? {this is multiplication}

Veronika [31]

Answer:

156.654

Step-by-step explanation:

162*.967=156.654

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2 years ago

Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector

Feliz [49]

$f(x,y,z)=2z^2x+y^3$

$f$ has gradient

$\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k$

which at the point (-1, 4, 3) has a value of

$\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k$

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say $\vec u=15\,\vec\imath+25\,\vec\jmath$ , in which case we have

$\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}$

Then the derivative of $f$ at (-1, 4, 3) in the direction of $\vec u$ is

$D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}$

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3 years ago