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Readme [11.4K]
3 years ago
9

HELP ASAPPP What are two ratios using the following information?There are 15 males teachers in the school and 35 female teachers

in the school. Please write both full answers below.
Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
8 0

Answer:

15:35,15 to 35

Step-by-step explanation:

3241004551 [841]3 years ago
3 0

Answer:

15 male teachers:35 female teachers

35 female teachers: 15 male teachers

Step-by-step explanation:

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If A+B = 76 & A -B = 38 What is A/B?
Igoryamba
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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,<em> s(s^2+s+1)</em> 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 <em>A</em>, <em>B</em> and <em>C</em> are unknown constants. The numerator of <em>s</em> is a constant <em>A</em>, because <em>s</em> is linear, the numerator of <em>s^2+s+1</em> is a linear expression <em>Bs+C</em> because <em>s^2+s+1</em> 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 <em>A=1</em>, <em>B=-1</em>, <em>C=0</em>. 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 <em>a=-1/2</em> and <em>b=(√3)/2</em>. 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}]

Finally:

\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)

7 0
4 years ago
I begg someone just please helppp
Temka [501]

Answer:

#4.

- 3 < - 2 1/2

If you consider the number line, the numbers increase from left to right. The number on the left is smaller than a number on the right.

- 3 is smaller than -2 1/2, it means -3 is on the left from the -2 1/2 on the number line.

#5.

Jada's score is positive, it means the number reflecting the score is to the right from zero on the number line.

Isabel's score is negative, it means the number reflecting the score is to the left from zero on the number line.

Since Isabel's score is to the left  from the Jada's score, she has lower score than Jada.

With the lowest score, Isabel is the winner.

6 0
3 years ago
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