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3 years ago

I need some help with this and a few others.

Mathematics

1 answer:

finlep [7]3 years ago

5 0

Answer:

x^3y^9

Step-by-step explanation:

Simplify both sides: Add powers of x together and powers of y together

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4) If g(x) = 13x2 what is the value of g(-3)?

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3 years ago

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Graph f(x)=|x−2|+3 .<br><br> Use the ray tool to graph the function.

maks197457 [2]

Answer:

The graph will be 2 units away from the origin on positive $x-axis$ and three units upward from the origin towards $y-axis$ .

Step-by-step explanation:

Here is a graph attached with it.

To graph $\left | x-2 \right |+3$ we know that positive $\left | x \right |$ is a $V$ shaped from the origin.

Key points:

To move rightward there must be a negative inside the parentheses.
And to move upward we must have positive $b = constant$ .

If we have to move towards $x-axis$ then we must have negative inside it.

And if we have to move upward in $y-axis$ positive we must have positive constant value.

So the graph will be three units upward and two units rightward with a V-shaped ray.

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4 years ago

In a class of 32 students, 16 students study Spanish, 14 students are in band, 6 students are in both Spanish and band. How many

Tanya [424]

Answer:

Step-by-step explanation:

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3 years ago

Use integration by parts to find the integrals in Exercise.<br> x^3 ln x dx.

34kurt

Answer:

$\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

Step-by-step explanation:

We have been given an indefinite integral $\int \:x^3\:ln\:x\:dx$ . We are asked to find the value of the integral using integration by parts.

$\int\: u\text{dv}=uv-\int\: v\text{du}$

Let $u=\text{ln}(x)$ , $v'=x^3$ .

Now, we will find du and v as shown below:

$\frac{du}{dx}=\frac{d}{dx}(\text{ln}(x))$

$\frac{du}{dx}=\frac{1}{x}$

$du=\frac{1}{x}dx$

$v=\frac{x^{3+1}}{3+1}=\frac{x^{4}}{4}$

Upon substituting our values in integration by parts formula, we will get:

$\int \:x^3\:\text{ln}\:(x)\:dx=\text{ln}(x)*\frac{x^4}{4}-\int\: \frac{x^4}{4}*\frac{1}{x}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\int\: \frac{x^3}{4}dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}\int\: x^3dx$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^{3+1}}{3+1}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{1}{4}*\frac{x^4}{4}+C$

$\int \:x^3\:\text{ln}\:(x)\:dx=\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$

Therefore, our required integral would be $\frac{\text{ln}(x)x^4}{4}-\frac{x^4}{16}+C$ .

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3 years ago

PLEASE HELP<br> Describe the graph of y=1/2x-10 -3 compared to the graph of y=1/x

jek_recluse [69]

Answer: The graphs are attached.

Step-by-step explanation: We are to describe the graph of the function $y=\dfrac{1}{2x-10}-3$ compared to the graph of the function $y=\dfrac{1}{x}.$

The graphs of both the functions are shown in the attached figure.

We can see that the graph of the function $y=\dfrac{1}{2x-10}-3$ is stretched by a factor of 0.5, shifted 5 units to the right and 3 units downwards as compared to the graph of the function $y=\dfrac{1}{x}.$

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3 years ago

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