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3 years ago

Evaluate each expression |-16| -|-1|

Mathematics

1 answer:

Maslowich3 years ago

4 0

Understanding the Absolute Value.

First, know what the absolute value is.

The absolute value is the value that determines how far the value is from 0.

For example, The absolute value of -5 is far from 0 5 units. Therefore the absolute value of -5 equals 5.

Basic Absolute Value Defines

| a | = a

- | a | = -a

| - a | = a

Back to the question. To evaluate those expressions, we use the defines of absolute value.

|-16| = 16

|-1| = 1

16-(1)

Then remove the brackets. 16 - 1 = 15

Therefore, the answer is 15.

The answer above is when being subtracted and evaluated from both 16-1

Evaluating for each expressions would be

|-16| = 16

-|-1| = -1

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3 years ago

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Step-by-step explanation:

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VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

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2 years ago

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Harlamova29_29 [7]

I typed right on the problem set.

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3 years ago

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djyliett [7]

Answer:

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3 years ago