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Simora [160]
3 years ago
15

Please please help me!!!!

Mathematics
1 answer:
exis [7]3 years ago
3 0
The answer is option 2
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Write an appropriate direct variation equation if y = 45 when x = 5.
Firdavs [7]
45×5 or 45÷5 i hope this helps if it dont im so so very very sorry for you
4 0
3 years ago
Which angle pairs are supplementary? Check all that apply. ∠1 and ∠2 ∠4 and ∠3 ∠4 and ∠5 ∠7 and ∠5 ∠3 and ∠6.
Alla [95]

Answer:

∠4 and ∠3

∠4 and ∠5

∠3 and ∠6.

Step-by-step explanation

Find the diagram attached

First you must note that the sum of two supplementary angles is 180°

Also, since the sum of angle on a straight line is 180°, the angles that are supplementary must lie on the same straight line.

Based on the diagram, the pair of angles that lies on the same straight line are ∠4 and ∠3, ∠4 and ∠5 and ∠3 and ∠6. If that is the case, hence this angles are supplementary.

6 0
2 years ago
Please help me , please
Tatiana [17]

Answer:

c is correct

Step-by-step explanation:

5 0
3 years ago
A=<br> f forr<br> 2<br> =<br> o<br> Need help with that plz help me
spayn [35]

Answer:

r = 2a + k

Step-by-step explanation:

Hope this helps :D

5 0
3 years ago
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
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