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3 years ago

Help with this thank you

Mathematics

1 answer:

dimulka [17.4K]3 years ago

6 0

Answer:

17 is the answer

Step-by-step explanation:

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Is 5/4 times 6, 11/8 times 6, and 3/8 times 6 products greater than 6?

weeeeeb [17]

Here were are dealing only with positive numbers and positive fractions whose numerators and denominators are positive integers.

If you multiply a number by 1, you get that number.

For example, 1 * 5 = 5; 1 * 8 = 8

If you multiply a number by a positive number less than one, the result is smaller than the number you started with.

For example, 1/2 * 6 = 3. Since 1/2 is less than 1, multiplying 1/2 by 6 gives a result smaller than 6. 3 is smaller than 6.

If you multiply a number by a number greater than 1, the result is greater than the number you started with.

For example, 2 * 6 = 12. Since you multiplied 6 by a number greater than 1, which 2 is, you get 12 which is greater than 6. 5/4 * 20 = 25. Since 20 was multiplied by 5/4, and 5/4 is greater than 1, the result, 25, is greater than 20.

It's easy to tell if a fraction is greater than or less than 1. If the numerator is less than the denominator, the fraction is less than 1.

Examples of fractions less then 1: 1/3, 1/6, 5/6, 11/12, 18/121. In every case to the left, the numerator is less than the denominator, so all fractions are lees than 1.

If the numerator is greater than the denominator, then the fraction is greater than 1.

Examples of fractions that are greater than 1: 4/3, 11/8, 3/2, 121/100, 17/18.

5/4 is greater than 1, so 5/4 * 6 is greater than 6.

11/8 is greater than 1, so 11/8 * 6 is greater than 6.

3/8 is less than 1, so 3/8 * 6 is less than 6.

7 0

3 years ago

Read 2 more answers

The bookstore sold m books on Monday and t books on Tuesday. But on Wednesday, r books were

Amiraneli [1.4K]

Answer:

Step-by-step explanation:

Label the books returned "b", and the books sold on monday "m", and the books sold on tuesday "t", and the books returned "r".

the formula:

$b = (m + t) - r$

(you can remove the brackets)

3 0

2 years ago

Sheila is a wildlife biologist. Her daily task is to count the number of wild turkeys and white-tail deer in a large game reserv

Lilit [14]

Given:
wild turkey ; 12 in the first hour and cumulative number increases by 40% per hour
white-tail deer ; 18 in the first hour and 10 deer each hour after that.

wild turkery: 12 * (1.40)^n-1 ; n is the number of hour
white-tail deer: 18 + 10^n-1 ; n is the number of hour

n wild turkey white-tail deer
1 12 18
2 17 28
3 24 38
4 33 48
5 46 58
6 65 68
7 90 78

It would be hour 7 after sunrise that the cumulative count of the turkeys first exceed the cumulative count of the deer.

4 0

3 years ago

How do I solve this ?

Alja [10]

Answer:

g(-9) = 1

Step-by-step explanation:

g(-9) is the value of the function at x = -9

Go over the x = -9 and go up until you hit the blue line

Read the value of y

g(-9) = 1

5 0

3 years ago

Read 2 more answers

Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x) = sin x, c = 3π/4

anyanavicka [17]

Answer:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Step-by-step explanation:

Given

$f(x) = \sin x\\$

$c = \frac{3\pi}{4}$

Required

Find the Taylor series

The Taylor series of a function is defines as:

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

We have:

$c = \frac{3\pi}{4}$

$f(x) = \sin x\\$

$f(c) = \sin(c)$

$f(c) = \sin(\frac{3\pi}{4})$

This gives:

$f(c) = \frac{1}{\sqrt 2}$

We have:

$f(c) = \sin(\frac{3\pi}{4})$

Differentiate

$f'(c) = \cos(\frac{3\pi}{4})$

This gives:

$f'(c) = -\frac{1}{\sqrt 2}$

We have:

$f'(c) = \cos(\frac{3\pi}{4})$

Differentiate

$f"(c) = -\sin(\frac{3\pi}{4})$

This gives:

$f"(c) = -\frac{1}{\sqrt 2}$

We have:

$f"(c) = -\sin(\frac{3\pi}{4})$

Differentiate

$f"'(c) = -\cos(\frac{3\pi}{4})$

This gives:

$f"'(c) = - * -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

So, we have:

$f(c) = \frac{1}{\sqrt 2}$

$f'(c) = -\frac{1}{\sqrt 2}$

$f"(c) = -\frac{1}{\sqrt 2}$

$f"'(c) = \frac{1}{\sqrt 2}$

$f(x) = f(c) + f'(c)(x -c) + \frac{f"(c)}{2!}(x-c)^2 + \frac{f"'(c)}{3!}(x-c)^3 + ........ + \frac{f*n(c)}{n!}(x-c)^n$

becomes

$f(x) = \frac{1}{\sqrt 2} - \frac{1}{\sqrt 2}(x - \frac{3\pi}{4}) -\frac{1/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{1/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Rewrite as:

$f(x) = \frac{1}{\sqrt 2} + \frac{(-1)}{\sqrt 2}(x - \frac{3\pi}{4}) +\frac{(-1)/\sqrt 2}{2!}(x - \frac{3\pi}{4})^2 +\frac{(-1)^2/\sqrt 2}{3!}(x - \frac{3\pi}{4})^3 + ... +\frac{f^n(c)}{n!}(x - \frac{3\pi}{4})^n$

Generally, the expression becomes

$f(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

Hence:

$\sin(x) = \sum\limit^{\infty}_{n = 0} \frac{1}{\sqrt 2}\frac{(-1)^{n(n+1)/2}}{n!}(x - \frac{3\pi}{4})^n$

3 0

2 years ago