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3 years ago

23 divided by x/3 =41

Mathematics

2 answers:

vlada-n [284]3 years ago

6 0

23=41x/3 ; 41x=69; x=69/41

Zepler [3.9K]3 years ago

3 0

Answer:314.3 repeating

Step-by-step explanation:

Multiply 41 by 23

Then decide by 3 to get x

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List the given numbers from least to greatest -5, 3.14, 1/4, -3/2, 49

seraphim [82]

-3/2,-5,1/4,3.14,49 hope that helped!

8 0

3 years ago

What can be best described as a relationship between two quantities in which one quantity is dependent upon the other?

Anastaziya [24]

Answer:

a function

Step-by-step explanation:

a function is where u plug something in and get another thing out. the entire function is dependent on another independent variable

5 0

4 years ago

I need the work to this please

wlad13 [49]

Answer:

? = 14

Step-by-step explanation:

If ∆EFG ~ ∆CBA, we have the proportion

FG/FE = BA/BC . . . . . any pairs of corresponding sides will have the same ratio for similar triangles. (It is convenient to put the unknown in the numerator.)

?/18 = 21/27 . . . . . . filling in the given numbers

Multiplying by 18, we have ...

? = 18·21/27

? = 14

_____

If the triangles are not designated as being similar, the problem is unworkable.

6 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

What do you do in your free time? :D

damaskus [11]

Answer:

sleep most definitely worth it

3 0

3 years ago

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