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4 years ago

Solve 5c − c + 10 = 34.

Mathematics

1 answer:

makkiz [27]4 years ago

3 0

Answer: 6

Step-by-step explanation: 1. 5c-c= 4c. The equation would then be 4c+10=34.

2. From there, you subtract 10 from both sides of the equation. By doing this, you have the variable and the nonvariable on separate sides of the equation.

3. After doing that, you should have 4c=24. To get the variable by itself, divide both sides by 4. 4c/4 is c and 24/4 is 6.

The final answer is 6. Hope this helped you:)

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4. Give 5 possible solutions to the equation: y = -2x + 5 X Y

barxatty [35]

Answer:

(0, 5), (1, 3), (2, 1), (3, -1), (4, -3)

Step-by-step explanation:

To find the possible solutions, we just plug in a random value for x and then solve for y. When x - 0, y = 5. When x = 1, y = 3. We just continue doing this until we have 5 different solutions.

7 0

2 years ago

The math department needs to buy new textbooks and laptops for the computer science classroom. The textbooks cost $116.00 each,

Oksana_A [137]

Answer:

6 laptops.

Step-by-step explanation:

$6500 is the budget. If each textbook costs $116, and 30 textbooks are needed, you multiply.

116 x 30 = 3480.

3480 was spent on the textbooks, and the rest needs to go into laptops.

6500 - 3480 = 3020.

Now to find out how many laptops can be bought with the remaining money, divide.

3020 / 439 = 6.88 (You can't buy a portion of a laptop, so you have to take the biggest whole number, which is 6).

6 0

4 years ago

Need help on these three. thanks:))

Anton [14]

1. 0.3% or 30%
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4 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:

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

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

Graph the line that passes through the two lines (1,5/2), (-1/2,-1/4)

mars1129 [50]

Answer:

Step-by-step explanation:

7 0

3 years ago