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3 years ago

Choose the correct simplification of the expression (-9a^3b^2c^10)^0

Mathematics

1 answer:

Flauer [41]3 years ago

8 0

Option C: 1 is the simplification of the expression $\left(-9 a^{3} b^{2} c^{10}\right)^{0}$

Explanation:

The given expression is $\left(-9 a^{3} b^{2} c^{10}\right)^{0}$

We need to simplify the given expression.

Simplification:

To simplify the given expression, let us apply the exponent rule, $a^{0}=1$ where $a \neq 0$

Thus, the the exponent rule $a^{0}=1$ means that any value (except zero) to the power of zero is equal to one.

Thus, the expression $\left(-9 a^{3} b^{2} c^{10}\right)^{0}$ becomes,

$\left(-9 a^{3} b^{2} c^{10}\right)^{0}=1$

Therefore, the simplified expression is 1

Hence, Option C is the correct answer.

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Based on the graph of the line below, which describes the relationship between the value of the y-intercept and the

stepan [7]

Answer:

y intercept = 3

$m = \frac{1}{3}$ --- slope

Step-by-step explanation:

Given

The attached graph

Required

Determine the slope and the y intercept

The $y\ intercept$ is at the point where $x = 0$ .

From the graph:

$y = 3;\ when\ x = 0$

Hence,

y intercept = 3

Next, we calculate the slope using:

$m = \frac{y_1 - y_2}{x_1 - x_2}$

Where:

$(x_1,y_1) = (0,3)$

$(x_2,y_2) = (9,6)$

So, we have:

$m = \frac{3 - 6}{0 - 9}$

$m = \frac{-3}{- 9}$

$m = \frac{3}{9}$

$m = \frac{1}{3}$

6 0

3 years ago

B-2-11. Find the inverse Laplace transform of s + 1/s(s^2 + s +1)

Aleksandr-060686 [28]

Answer:

$\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=1-e^{-t/2}cos(\frac{\sqrt{3} }{2}t )+\frac{e^{-t/2}}{\sqrt{3} }sin(\frac{\sqrt{3} }{2}t)$

Step-by-step explanation:

let's start by separating the fraction into two new smaller fractions

First, s(s^2+s+1) must be factorized the most, and it is already. Every factor will become the denominator of a new fraction.

$\frac{s+1}{s(s^{2} + s +1)}=\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}$

Where A, B and C are unknown constants. The numerator of s is a constant A, because s is linear, the numerator of s^2+s+1 is a linear expression Bs+C because s^2+s+1 is a quadratic expression.

Multiply both sides by the complete denominator:

$[{s(s^{2} + s +1)]\frac{s+1}{s(s^{2} + s +1)}=[\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}][{s(s^{2} + s +1)]$

Simplify, reorganize and compare every coefficient both sides:

$s+1=A(s^2 + s +1)+(Bs+C)(s)\\\\s+1=As^{2}+As+A+Bs^{2}+Cs\\\\0s^{2}+1s^{1}+1s^{0}=(A+B)s^{2}+(A+C)s^{1}+As^{0}\\\\0=A+B\\1=A+C\\1=A$

Solving the system, we find A=1, B=-1, C=0. Now:

$\frac{s+1}{s(s^{2} + s +1)}=\frac{1}{s}+\frac{-1s+0}{s^{2}+s+1}=\frac{1}{s}-\frac{s}{s^{2}+s+1}$

Then, we can solve the inverse Laplace transform with simplified expressions:

$\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{\frac{1}{s}\}-\mathcal{L}^{-1}\{\frac{s}{s^{2}+s+1}\}$

The first inverse Laplace transform has the formula:

$\mathcal{L}^{-1}\{\frac{A}{s}\}=A\\ \\\mathcal{L}^{-1}\{\frac{1}{s}\}=1\\$

For:

$\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}$

We have the formulas:

$\mathcal{L}^{-1}\{\frac{s-a}{(s-a)^{2}+b^{2}}\}=e^{at}cos(bt)\\\\\mathcal{L}^{-1}\{\frac{b}{(s-a)^{2}+b^{2}}\}=e^{at}sin(bt)$

We have to factorize the denominator:

$-\frac{s}{s^{2}+s+1}=-\frac{s+1/2-1/2}{(s+1/2)^{2}+3/4}=-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}$

It means that:

$\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}\}$

$\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}+\mathcal{L}^{-1}\{\frac{1/2}{(s+1/2)^{2}+3/4}\}\\\\\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}+\frac{1}{2} \mathcal{L}^{-1}\{\frac{1}{(s+1/2)^{2}+3/4}\}$

So a=-1/2 and b=(√3)/2. Then:

$\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}=e^{-\frac{t}{2}}[cos\frac{\sqrt{3}t }{2}]\\\\\\\frac{1}{2}[\frac{2}{\sqrt{3} } ]\mathcal{L}^{-1}\{\frac{\sqrt{3}/2 }{(s+1/2)^{2}+3/4}\}=\frac{1}{\sqrt{3} } e^{-\frac{t}{2}}[sin\frac{\sqrt{3}t }{2}]$