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3 years ago

Help me with these simple math problems

Mathematics

1 answer:

earnstyle [38]3 years ago

8 0

Answer:

Please read the answer below.

Step-by-step explanation:

1. Australia:

75 * 1.87 =140 Australian dollars

2. Brazil:

75 * 2.32 = 174 Reals

3. Britain:

75 * 0.69 = 52 Pounds

4. Canada:

75 * 1.60 = 120 Canadian dollars

5. China:

75 * 8.28 = 621 Yuan

6. Denmark:

75 * 8.43 = 632 Kroner

7. Japan:

75 * 131.55 = 9,866 Yen

8. Mexico:

75 * 9.19 = 689 Mexican pesos

9. South Africa:

75 * 11.9 = 893 Rands

10. Sweden:

75 * 10.61 = 796 Kronor

11. Switzerland:

75 * 1.68 = 126 Francs

12. Thailand:

75 * 44.18 = 3,314 Baht

All currencies rounded to the next integer.

Note: Same answer to question 14454918, answered by me today.

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Which of the following is the equation of the function below?

puteri [66]

Answer:

The equation of the graph below is y = 0.5 csc[0.5(x + π/2)] - 1 ⇒ 3rd answer

Step-by-step explanation:

* Lets revise the trigonometry transformation

- If the equation is y = A csc(B(x + C)) + D

# Amplitude is A

- The Amplitude is the height from the center line to the peak .

Or we can measure the height from highest to lowest points and

divide that by 2

# period is 2π/B

- The Period goes from one peak to the next

# phase shift is C (positive is to the left)

- The Phase Shift is how far the function is shifted horizontally

from the usual position.

# vertical shift is D

- The Vertical Shift is how far the function is shifted vertically from

the usual position.

- If y = csc(x)

∴ A = 1 , B = 1 , C = 0 , D = 0

- That means the amplitude is 1, the period is 2π, no phase shift

or vertical shift

* Now lets solve the problem

- From the graph

# The amplitude = (-0.5 - -1.5)/2 = 0.5

∴ A = 0.5

# The period is from 2.5π to -1.5π

∴ The period is 4π

∵ The period = 2π/B

∴ 4π = 2π/B ⇒ by cross multiplication

∴ B = 2π/4π = 1/2 = 0.5

* There is only one answer has A = 0.5 and B = 0.5

∴ y = 0.5 csc[0.5(x + π/2)] - 1

* The equation of the graph below is 0.5 csc[0.5(x + π/2)] - 1

* For more understand look to the color graph

- The blue graph is y = csc(x)

- The green graph is y = 0.5 csc[0.5(x + π/2)] - 1

5 0

2 years ago

Read 2 more answers

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

To find the first three non-zero terms you need to replace n=3 into the sum

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$