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pychu [463]
2 years ago
13

(9a^2 + 7a + 8) - (2a^2 + 5a + 1)

Mathematics
1 answer:
Sergio [31]2 years ago
8 0

Answer:  7a^2+2a+7

Step-by-step explanation:Distribute the Negative Sign:

=9a2+7a+8+−1(2a2+5a+1)

=9a2+7a+8+−1(2a2)+−1(5a)+(−1)(1)

=9a2+7a+8+−2a2+−5a+−1

Combine Like Terms:

=9a2+7a+8+−2a2+−5a+−1

=(9a2+−2a2)+(7a+−5a)+(8+−1)

=7a2+2a+7

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If GEOM is a square, find m∠GQE.
Alexxandr [17]

Answer:

i think it is 180

Step-by-step explanation:

because 180 is the total degrees of a triangle

7 0
3 years ago
Find the circumference of the circle in terms of π.
SSSSS [86.1K]

Answer:

  • C) 12π cm

Step-by-step explanation:

<u>Circumference formula:</u>

  • C = 2πr

Given r = 6 cm

<u>Circumference is:</u>

  • C = 2π*6 cm = 12π cm

Correct choice is C

4 0
2 years ago
You randomly draw twice from this deck of cards
Serggg [28]

The probability of not drawing C in neither draw is P = 0.5

<h3>How to get the probability?</h3>

All the cards have the same probability of being drawn, in this case, our set of cards is {F, D, C, G}

The probability of not drawing C is equal to the probability of drawing F, D or G. So we have 3 options out of 4, then the probability is:

p = 3/4.

Now we draw another, this time there are 3 cards, one of these is C, and the other two cards are not C. Then the probability of not drawing C again is equal to 2 over 3.

q = 2/3.

The joint probability (for both of these events to happen) is equal to the product of the individual probabilities:

P = p*q = (3/4)*(2/3) = 0.5

If you want to learn more about probability, you can read:

brainly.com/question/251701

3 0
2 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
Four friends each bought a movie ticket and two boxes of candy at a movie theater. The tickets were $ 7.50 and the boxes of cand
777dan777 [17]
Subtract the cost of tickets from the total. 4x7.5=30. 58-30=28. Eight boxes of candy were purchased in total. Use c to represent candy. 8c=28. Finally, divide both by 8. 28/8= 3.5. The answer is $3.50.
6 0
3 years ago
Read 2 more answers
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