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user100 [1]
4 years ago
13

One to $255 by working 20 hours each week writing equation to find out how much Lauren earns per hour use H to represent the hou

rly rate so I'll be equation to find laws hourly rate
show your work?
Mathematics
2 answers:
Taya2010 [7]4 years ago
4 0
255/20=12.75 should be the answer to you question
Mashutka [201]4 years ago
4 0
Since Lauren earns $255 for working 20 hours; we'll have to divide that to see what she gets per hour. We divide 255 by 20 (255/20) which would give us 12.75, so she get $12.75 per hour. Now that we've done that we can now start creating the formula. X will stand for how much money she earns in total (X=). We will now put 12.75h. Now we know that 12.75 is supposed to be multiplied by how many hours she works (h). After that we're done! I hope this helps ^^ Be sure to give me brainliest answer if it did help.
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One positive integer is 7
laiz [17]
You use a system of equations to solve this.
x will be one integer and y will be the other.

2y - 7 = x
x^2 + y^2 = 433

You can plug (2y - 7) in for x in the second equation.

(2y - 7)^2 + y^2 = 433
4y^2 - 28y + 49 + y^2 = 433
5y^2 - 28y -384 = 0
y = 12 or -32/5

It has to be 12 since -32/5 is not an integer.

Plug 12 in for y to get x.

2(12) - 7 = x
x = 17

The integers are 12 and 17.
6 0
4 years ago
37 (x−5) + 2 = 2 (3−2x)<br><br> Solve for x
jenyasd209 [6]
37(x-5) + 2 = 2(3-2x)...distribute thru the parenthesis
37x - 185 + 2 = 6 - 4x
37x - 183 = 6 - 4x
37x + 4x = 6 + 183
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7 0
3 years ago
What term describes the monomial 14xyz?<br> constant<br> linear<br> quadratic<br> cubic
damaskus [11]

Answer:

Cubic

Step-by-step explanation:

14xyz is a cubic polynomial

5 0
3 years ago
HELPPPP PLEASE!!!! Don't understand
iragen [17]

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7 0
3 years ago
Read 2 more answers
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
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