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2 years ago

(6x + 1) 37 Solve for x.

Mathematics

2 answers:

____ [38]2 years ago

7 0

Answer:

atleast 5

Step-by-step explanation:

antoniya [11.8K]2 years ago

7 0

Answer:

this is your ans mark brainliest pls

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Pls help solve this...No hacker links

Cerrena [4.2K]

Answer:

x=13

Step-by-step explanation:

18-5=13. You have to subtract 5 from 5 itself and then 18.

8 0

2 years ago

Read 2 more answers

Y=2x-5 3x+8y+32=56 What does y and x equal?

AVprozaik [17]

Y = 2x - 5
3x + 8y + 32 = 56
3x + 8(2x - 5) + 32 = 56
3x + 8(2x) - 8(5) + 32 = 56
3x + 16x - 40 + 32 = 56
19x - 8 = 56
 + 8 + 8
 19x = 64
 19 19
 x = 3.4
y = 2(3.4) - 5
y = 6.8 - 5
y = 1.8
(x, y) = (3.4, 1.8)

5 0

2 years ago

A markdown that is less than 1%

Alexxx [7]

Then mark it down by less then 1%, of by whatever the number would be.

8 0

2 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

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

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

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

7 0

2 years ago

Evaluate the expression for the given values. 8x+2yz , where x=12 , y = 4, and z = 2 Enter your answer in the box. need this ASA

FromTheMoon [43]

112 is the answer I think

3 0

3 years ago

Read 2 more answers