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Angelina_Jolie [31]
3 years ago
11

R + 11 + 8r= 29 can you help me

Mathematics
2 answers:
Serggg [28]3 years ago
6 0

Answer:

9

Step-by-step explanation:

r+8r+11=29

9r+11=29

9r=29-11

9r=18

r=18-9

r=9

<em><u>HOPE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>HELPS</u></em><em><u> </u></em><em><u>YOU</u></em><em><u> </u></em>

<em><u>STAY</u></em><em><u> </u></em><em><u>SAFE</u></em><em><u> </u></em><em><u>AND</u></em><em><u> </u></em><em><u>ENJOY</u></em><em><u> </u></em><em><u>YOUR</u></em><em><u> </u></em><em><u>DAY</u></em><em><u> </u></em>

poizon [28]3 years ago
5 0

<em>hello </em><em>there </em><em>I </em><em>can </em><em>help </em><em>you</em>

<em>=  \:  \:  \:  \: r = 2</em>

<em>hope </em><em>it</em><em> helps</em>

<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on</em><em> learning</em>

<em>mark </em><em>me</em><em> as</em><em> brainlist</em><em> plss</em>

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What is the measure of the unknown angle?
Romashka-Z-Leto [24]

Answer:

63

Step-by-step explanation:

a entire circle is 360⁰

so you do:

360-297

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Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.
Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Step-by-step explanation:

GIven that:

f(x) = 5e^{-x^2} cos (4x)

The Maclaurin series of cos x can be expressed as :

\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...  \ \ \ (1)}

\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0}  \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!}  -\dfrac{x^6}{3!}+... \ \ \  (2)}

From equation(1), substituting x with (4x), Then:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}

The first three terms of cos (4x) is:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}

\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}

\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}

Multiplying equation (2) with (3); we have :

\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }

\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Finally , multiplying 5 with \mathtt{ e^{-x^2} cos (4x) } ; we have:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

7 0
3 years ago
What expression is equivalent to 8/9 divided 3/4
Semmy [17]

Answer:

32/27 or 1/5/27 or a repeating decimal 1.185

Step-by-step explanation:

4 0
3 years ago
Expand &amp; simplify
Klio2033 [76]

Answer:

24x-12

Step-by-step explanation:

You must distribute first (8x+4+16x-16) Then comine like terms (8x+16x) (4-16). (24x-12)

Hope this helped!

5 0
1 year ago
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