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3 years ago

PLEASE HELP MEEEE

Mathematics

1 answer:

Rus_ich [418]3 years ago

4 0

Answer:

15771 > - 7

Step-by-step explanation:

Comparing the elevation of the two points ;

Point with elevation of 15771 is positive meaning it occurs above sea level and this higher than the point with an elevation of - 7 which is occurring below sea level

Hence, to compare both elevations :

Montblanc has higher elevation than Rhone river delta

Montblanc elevation > Rhone river delta

15771 > - 7

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Ulleksa [173]

Step-by-step explanation:

are you having problems with something if so message

4 0

2 years ago

Solve the following inequality in two ways and make a claim as to which

NikAS [45]

Answer:

The solution of the given inequality n < 5

Step-by-step explanation:

Explanation:-

Given inequality

$\frac{1}{4} (8 n -12 ) < n +2$

Step(i):-

Cross multiplication '4' we get

8 n - 12 < 4 ( n +2)

8 n - 12 < 4 n + 4×2

8 n - 12 < 4 n +8

Subtracting '4 n ' on both sides, we get

8 n - 4 n - 12 < 4 n - 4 n +8

4 n - 12 < 8

Step(ii):-

Adding ' 12 ' on both sides , we get

4 n -12 + 12 < 8 + 12

4 n < 20

Dividing "4' on both sides , we get

n < 5

The solution is n < 5

Conclusion:-

The solution of the given inequality n < 5

5 0

3 years ago

PLEASE PLEASE PLZZ HELP :(((((

xxMikexx [17]

Answer:

Step-by-step explanation:

Also the first one ends in divided by 1/4

4 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

HELP NEED ASAP FOR TOMORROW ITS FOR MY HOMEWORK!! ILL MARK BRAINLIEST IF RIGHT!!

Slav-nsk [51]

Answer:

y ≥ -x +2

Step-by-step explanation:

The solid line has a slope of -1 and a y-intercept of 2, so its equation in slope-intercept form is ...

y = -x +2

The shaded area is above this line, and the line is part of the solution set, so we want an inequality that has "y" and the comparison symbol in this order: "y ≥" or "≤ y".

We already have an equation with "y" on the left, above, so we just need to introduce the comparison symbol:

y ≥ -x +2

Another way to write this is ...

x + y ≥ 2

4 0

3 years ago