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2 years ago

How to solve 15x^3/4

Mathematics

2 answers:

valentinak56 [21]2 years ago

8 0

Answer:

Well...without the x I know it is 843.75 but I don't know the fraction of it so sorry! I try my best to help people!

Elena L [17]2 years ago

3 0

Answer:

Simplify —

Equation at the end of step

15x • —

STEP

Final result :

45x

———

Step-by-step explanation:

You might be interested in

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

2 years ago

Someone please help. Thank u so much

Nonamiya [84]

1. $120.66+15.55=$136.21
2. $700.00-$136.21=?
3. The answer is in the 1st and 2nd question

7 0

3 years ago

To check a solution, you can ____________the solutions into the equations and verify that both equations are true.

kondor19780726 [428]

Answer: To check a solution, you can substitute the solutions into the equations and verify that both equations are true.

Step-by-step explanation:

7 0

2 years ago

James has $32 and earns $10 per week for his allowance. What is the initial value for the scenario described?

nignag [31]

32 is your initial value, because that is what he starts with

6 0

2 years ago

Read 2 more answers

I really need help with 15 and 16 please

Mandarinka [93]

Answer:

15. 50 kg

16a. 5 kg

16b. 3.75 kg

Step-by-step explanation:

The formula relating force, mass, and acceleration can be solved for mass. This formula will apply to both problems. We'll use m for both "mass" and "meters". We presume you can avoid getting mixed up by understanding that meters is used in the context of acceleration: m/s².

F = ma

m = F/a . . . . . divide by a

15. m = (250 N)/(5 m/s²) = 50 kg

16a. m = (15 N)/(3 m/s²) = 5 kg

16b. m = (15 N)/(4 m/s²) = 3.75 kg

_____

<em>Comment on units</em>

Especially for physics problems, I like to keep the units with the numbers. It is helpful to remember that Newtons are equivalent to kg·m/s². So, dividing Newtons by acceleration in m/s² will give mass in kg. Since you're familiar with F=ma, it's not too hard to remember that the units of force (N) are the product of the units of mass (kg) and acceleration (m/s²).

6 0

3 years ago