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.
37(x-5) + 2 = 2(3-2x)...distribute thru the parenthesis
37x - 185 + 2 = 6 - 4x
37x - 183 = 6 - 4x
37x + 4x = 6 + 183
41x = 189
x = 189/41
x = 4.6
Answer:
Cubic
Step-by-step explanation:
14xyz is a cubic polynomial
Answer: 20% is your answer
Step-by-step explanation:
180-150=30
30 divided by 150 = .2 which is 20%
Answer:

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.

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)]](https://tex.z-dn.net/?f=%5B%7Bs%28s%5E%7B2%7D%20%2B%20s%20%2B1%29%5D%5Cfrac%7Bs%2B1%7D%7Bs%28s%5E%7B2%7D%20%2B%20s%20%2B1%29%7D%3D%5B%5Cfrac%7BA%7D%7Bs%7D%2B%5Cfrac%7BBs%2BC%7D%7Bs%5E%7B2%7D%2Bs%2B1%7D%5D%5B%7Bs%28s%5E%7B2%7D%20%2B%20s%20%2B1%29%5D)
Simplify, reorganize and compare every coefficient both sides:

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

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

The first inverse Laplace transform has the formula:

For:

We have the formulas:

We have to factorize the denominator:

It means that:


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}]](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5E%7B-1%7D%5C%7B-%5Cfrac%7Bs%2B1%2F2%7D%7B%28s%2B1%2F2%29%5E%7B2%7D%2B3%2F4%7D%5C%7D%3De%5E%7B-%5Cfrac%7Bt%7D%7B2%7D%7D%5Bcos%5Cfrac%7B%5Csqrt%7B3%7Dt%20%7D%7B2%7D%5D%5C%5C%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%5B%5Cfrac%7B2%7D%7B%5Csqrt%7B3%7D%20%7D%20%5D%5Cmathcal%7BL%7D%5E%7B-1%7D%5C%7B%5Cfrac%7B%5Csqrt%7B3%7D%2F2%20%7D%7B%28s%2B1%2F2%29%5E%7B2%7D%2B3%2F4%7D%5C%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%20%7D%20e%5E%7B-%5Cfrac%7Bt%7D%7B2%7D%7D%5Bsin%5Cfrac%7B%5Csqrt%7B3%7Dt%20%7D%7B2%7D%5D)
Finally:
