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3 years ago

Sonji plots -9 on a number line to represent the depth of her sailboat underwater and labels it B. She plots positive 20

Mathematics

1 answer:

AfilCa [17]3 years ago

8 0

Answer: the answer ould be b hope this helps could you brainliest plz

Step-by-step explanation:

well bc one of the number on the cordinate plane is a negative and the other is a positive than it would be answer choice b hope this helps and could you PLZ brainliest please

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PLEASE ANSWER QUICKLY ASAP <br>READ THE QUESTION CAREFULLY PLEASE

Anna11 [10]

Answer:

140°

Step-by-step explanation:

total angle of a hexagon is 720. We know already 510°. That leaves 210°. The ratio between the two angles is 2:1, therefore angle CDE is 140° and angle DEF is 70 °

6 0

3 years ago

Read 2 more answers

Which is greater 2 2/3, 2.45, 2 2/5

timofeeve [1]

This is least to greatest

2 2/5 or 2.40

2.45

2 2/3

2 2/5 , 2.45 , 2 2/3

7 0

3 years ago

Read 2 more answers

Please help me and thank you:)

sashaice [31]

Okay. For these types of problems, you must do order of operations (PEMDAS). Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction. Mind you that you do these steps from left to right, and multiplication and division is done from left to right. Same thing with addition and subtraction. With that being said, here are your answers if you do the expressions correctly.

1. 12

2. 106

3. 42

4 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

An article in American Demographics claims that more than twice as many shoppers are out shopping on the weekends that during th

Monica [59]

Answer:

0.3085,0.2417,0.0045

Step-by-step explanation:

Given that X, the amount of money spent at shopping centers between 4 P.M. and 6 P.M. on Sundays has a normal distribution with mean $85 and with a standard deviation of $20.

X is N(85, 20)

To convert into std normal variate we use the following formula

$Z=\frac{x-85}{20}$