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2 years ago

I appreciate the help please

Mathematics

2 answers:

Hitman42 [59]2 years ago

8 0

Answer:

X = 37

Explanation:

Hope this helps!

8_murik_8 [283]2 years ago

5 0

Tough luck fool i aint giving u no answer u fool

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6−3(5x+2)−10x please explain the steps on how to simplify this polynomials.

kati45 [8]

Step-by-step explanation:

6-15x-6-10x

put the like terms together and it's now -15x-10x+6-6

-25+6-6

=-25

6 0

1 year ago

(WILL GIVE BRINALIST TO BEST ANWER) WHATS the measure of < 3

malfutka [58]

Answer:

∠3 = 60°

Step-by-step explanation:

Since g and h are parallel lines then

∠1 and ∠2 are same side interior angles and are supplementary, hence

4x + 36 +3x - 3 = 180

7x + 33 = 180 ( subtract 33 from both sides )

7x = 147 ( divide both sides by 7 )

x = 21

Thus ∠2 = (3 × 21) - 3 = 63 - 3 = 60°

∠ 2 and ∠3 are alternate angles and congruent, hence

∠3 = 60°

8 0

2 years ago

Estimate the quotient when 29,376 is divided by 29

Mnenie [13.5K]

The Quotient would be 1012 so I would just put 1000 if your supposed to estimate.

5 0

3 years ago

Read 2 more answers

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

2 years ago

This one too please!!!!!!!!!

Vinvika [58]

X= 0 and 1 are the answers

7 0

3 years ago