Thank to whoever answers this

Mathematics

2 answers:

blagie [28]3 years ago

8 0

I’m pretty sure the answer is A, i hope this helps!!

Xelga [282]3 years ago

3 0

Answer:

Hi! Do you need any help with anything?

Step-by-step explanation:

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Your food bill at a restaurant was 210 you leave a 15% tip what was total cost of your meal

s344n2d4d5 [400]

$241.50 due to my calculation

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4 years ago

Read 2 more answers

What is 1/2(4x+14)=2(x-7)

GREYUIT [131]

Answer:

no solution

Step-by-step explanation:

1/2(4x+14)=2(x-7)

Multiply both sides by 2.

4x + 14 = 4(x - 7)

4x + 14 = 4x - 28

Subtract 4x from both sides.

14 = -28

Since 14 = -28 is a false statement, there is no solution for this equation.

Answer: no solution

8 0

3 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$