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3 years ago

Help please I dont know what to do

Mathematics

2 answers:

KIM [24]3 years ago

7 0

Hope this helps:) the fracción is the 1/2 part

ser-zykov [4K]3 years ago

3 0

Answer:

7 1/2 = 1/2

15 = 1

22 1/2 = 1 1/2

30= 2

37 1/2= 2 1/2

Explanation: add 7 1/2 twice which equals 15 for 1 30 for two but have of 15 is 7 1/2 so add 15 and 7 1/2 to the other 2.

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Write in exponential form.<br><br>a · a · a · a · b · b · b · b · b · b

Nimfa-mama [501]

Answer:

A^4xb^6

Step-by-step explanation:

5 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

She works as a waitress at a local Emerald Monday restaurant. Her regular hourly wage is $15.50. She regularly works 40 hours pe

lina2011 [118]

Answer:

Her regular weekly pay is $62 and her annual salary is $3100.

Step-by-step explanation:

Given that her hourly wage=$15.50 / hour

As she works 40 hours per week, so her weekly pay = 15.50 x 40 = $62/ week.

If she works 50 weeks each year, so her yearly pay = 62x50=$3100 /year.

Hence, her regular weekly pay is $62 and her annual salary is $3100.

8 0

3 years ago

Describe the error made in subtracting the two

oksian1 [2.3K]

Answer:

You must subtract all terms of the numerator being subtracted out, not just the first term.

The negative 2 should have been subtracted out to get a numerator of x+1-x+2

The simplified numerator of the difference should be 3, not –1.

5 0

4 years ago

If two angles are both vertical and supplementary, can we determine the angles? Is it possible to be both vertical and complemen

Mars2501 [29]

Answer:

Two angles can be vertical and supplementary only if they add up to 90 degrees because there perpendicular.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers