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3 years ago

How to put 7/13 in simplest form

Mathematics

2 answers:

Shalnov [3]3 years ago

7 0

Answer:

cannot be reduced further.

Step-by-step explanation:

This is already is simplest form because 7 and 13 do not have any more factors so they cannot be broken down further. They are prime numbers.

Amiraneli [1.4K]3 years ago

5 0

Answer: 7/13 is already in simplest form.

Step-by-step explanation:

The sides cannot both be divided by the same number, meaning it is already in simplest form.

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tankabanditka [31]

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What are the solutions of the equation 2x2 = 2? Use a graph of the related function. A. There are two solutions: -1 and 1. B. Th

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7 0

3 years ago

Read 2 more answers

Let f(x) = 5x + 12. Find f−1(x).

krok68 [10]

Replace,

x = y

Or

x = f(x)

then us stay:

x = 5y + 12

Isolate 5y

5y = x - 12

Dividing both the sides by 5

y = x/5 - 12/5

Then, replace y = f^-(x)

f^-(x) = x/5 - 12/5

Hope this helps

7 0

3 years ago

Dy/dx if y = Ln (2x3 + 3x).

NeTakaya

Answer:

$\frac{6x^2+3}{2x^3+3x}$

Step-by-step explanation:

You need to apply the chain rule here.

There are few other requirements:

You will need to know how to differentiate $\ln(u)$ .

You will need to know how to differentiate polynomials as well.

So here are some rules we will be applying:

Assume $u=u(x) \text{ and } v=v(x)$

$\frac{d}{dx}\ln(u)=\frac{1}{u} \cdot \frac{du}{dx}$

$\text{ power rule } \frac{d}{dx}x^n=nx^{n-1}$

$\text{ constant multiply rule } \frac{d}{dx}c\cdot u=c \cdot \frac{du}{dx}$

$\text{ sum/difference rule } \frac{d}{dx}(u \pm v)=\frac{du}{dx} \pm \frac{dv}{dx}$

Those appear to be really all we need.

Let's do it:

$\frac{d}{dx}\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot \frac{d}{dx}(2x^3+3x)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (\frac{d}{dx}(2x^3)+\frac{d}{dx}(3x))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot \frac{dx^3}{dx}+3 \cdot \frac{dx}{dx})$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot 3x^2+3(1))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (6x^2+3)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{6x^2+3}{2x^3+3x}$

I tried to be very clear of how I used the rules I mentioned but all you have to do for derivative of natural log is derivative of inside over the inside.

Your answer is $\frac{dy}{dx}=\frac{(2x^3+3x)'}{2x^3+3x}=\frac{6x^2+3}{2x^3+3x}$ .

3 0

3 years ago