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3 years ago

Can you help me plz and think you

Mathematics

2 answers:

Lapatulllka [165]3 years ago

6 0

Answer:

m + 24

Step-by-step explanation:

6- (4 ÷ 2 × 5)? 4÷2=2 hope that this helps

lbvjy [14]3 years ago

3 0

Answer:

M -24 because math is a thing and I have to have a minimum of twenty characters

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Please help! 20 points if u answer

umka2103 [35]

Answer:

find the measure of J

180-45=135

find the measure of K

360-135=225

6 0

2 years ago

A 20 ounce box of cereal costs $3.20. A 12 ounce box of cereal costs $1.80. Which box of cereal is the better buy?

lisov135 [29]

Use Socratic it always works usually but it’s faster

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

4 years ago

3(x+4)-1=-7 plz help

tatiyna

Answer:

x = -6

Step-by-step explanation:

3(x+4)-1=-7

Add 1 to each side

3(x+4)-1+1=-7+1

3(x+4)=-6

Divide by 3

3/3(x+4)=-6/3

x+4 = -2

Subtract 4 from each side

x+4-4 = -2-4

x = -6

5 0

3 years ago

Read 2 more answers

Write an equation to find the nth term. Then find ₐ₂₄<br> 1, 3, 5, 7,.......

kirill115 [55]

The equation to find the nth term is:

an = a+(n-1)d

Where a is the first term in the sequence, which is 1 and d is the difference, which in this sequence the next term is found by adding 2 to the previous term, so d = 2 and n is the term you want to find.

The equation is:

an = 1 + (n-1)2

a24 = 1 + (24-1)2 = 1 + 23(2) = 1+46 = 47

5 0

3 years ago