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mixer [17]
3 years ago
5

1 Solve this quadratic using two different methods. Show both methods. x2 + 4x - 5 = 0

Mathematics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

x = -5 or x = 1

Step-by-step explanation:

x^{2} + 4x - 5 = 0\\(x+5)(x-1)=0\\

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Solve for X<br><br> PLEASE HELP ME
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Answer:

x = 10°

Step-by-step explanation:

Step 1:

92° + x° + 78° = 180°             Supplementary Angles

Step 2:

x° + 170° = 180°         Combine Like Terms

Step 3:

x° = 180° - 170°       Subtract 170° on both sides

Answer:

x = 10°

Hope This Helps :)

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the two cycles have center G where radius GI= 4ft, radius MG =6.5ft and MGC=72. Determine the total cost be to paint area MCAI i
Solnce55 [7]
I have no clue what the answer would be, I’ve worked on it for 25 minutes, I will get back to you if I figure it out :,)
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16 + 36 Drag numbers to the lines to create an equivalent expression in factored form
Goshia [24]
Answer:
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this is most likely the correct answer
7 0
3 years ago
Read 2 more answers
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
How did you get your answer, please show me your work
ivolga24 [154]
12) (shaded region) 12x25 =300
( un- shaded region) 6x10=60

300-60=240

240 is the answer for #12
6 0
3 years ago
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