3x + 6 = 12
First operation was subtraction (subtracting 6 to leave 3x alone)
Second operation was division (Dividing 3 from 3x to isolare x)
3x + 6 = 12
3x + 6 -(6) = 12 - 6
3x = 6
![\frac{3x}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B3x%7D%7B3%7D%20)
=
![\frac{6}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B6%7D%7B3%7D%20)
3 and 3 cancels out
x = 2
Answer: RR or Rr
A red allele present on at least one
of two homologous chromosomes.
Explaination:
Since red is dominant, only
one chromosome needs a red allele in
order for the plant to have red flowers.
Answer:7.05
Step-by-step explanation:
Answer:
aaannnd?
Step-by-step explanation:
Answer:
![\displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=-4 \ln |x|+5 \ln|x-1|+\dfrac{3}{x-1}+\text{C}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bx%5E2-4%7D%7Bx%5E3-2x%5E2%2Bx%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%3D-4%20%5Cln%20%7Cx%7C%2B5%20%5Cln%7Cx-1%7C%2B%5Cdfrac%7B3%7D%7Bx-1%7D%2B%5Ctext%7BC%7D)
Step-by-step explanation:
<u>Fundamental Theorem of Calculus</u>
![\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Ctext%7Bf%7D%28x%29%5C%3A%5Ctext%7Bd%7Dx%3D%5Ctext%7BF%7D%28x%29%2B%5Ctext%7BC%7D%20%5Ciff%20%5Ctext%7Bf%7D%28x%29%3D%5Cdfrac%7B%5Ctext%7Bd%7D%7D%7B%5Ctext%7Bd%7Dx%7D%28%5Ctext%7BF%7D%28x%29%29)
If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a <u>constant of integration</u>.
<u>Given integral</u>:
![\displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bx%5E2-4%7D%7Bx%5E3-2x%5E2%2Bx%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
Factor the denominator:
![\begin{aligned}\implies x^3-2x^2+x & = x(x^2-2x+1)\\& = x(x^2-x-x+1)\\& = x(x(x-1)-1(x-1))\\ & = x((x-1)(x-1))\\& = x(x-1)^2\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%20x%5E3-2x%5E2%2Bx%20%26%20%3D%20x%28x%5E2-2x%2B1%29%5C%5C%26%20%3D%20x%28x%5E2-x-x%2B1%29%5C%5C%26%20%3D%20x%28x%28x-1%29-1%28x-1%29%29%5C%5C%20%26%20%3D%20x%28%28x-1%29%28x-1%29%29%5C%5C%26%20%3D%20x%28x-1%29%5E2%5Cend%7Baligned%7D)
![\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int \dfrac{x^2-4}{x(x-1)^2}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bx%5E2-4%7D%7Bx%5E3-2x%5E2%2Bx%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%3D%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bx%5E2-4%7D%7Bx%28x-1%29%5E2%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
Take partial fractions of the given fraction by writing out the fraction as an identity:
![\begin{aligned}\dfrac{x^2-4}{x(x-1)^2} & \equiv \dfrac{A}{x}+\dfrac{B}{(x-1)}+\dfrac{C}{(x-1)^2}\\\\ \implies \dfrac{x^2-4}{x(x-1)^2} & \equiv \dfrac{A(x-1)^2}{x(x-1)^2}+\dfrac{Bx(x-1)}{x(x-1)^2}+\dfrac{Cx}{x(x-1)^2}\\\\ \implies x^2-4 & \equiv A(x-1)^2+Bx(x-1)+Cx \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdfrac%7Bx%5E2-4%7D%7Bx%28x-1%29%5E2%7D%20%26%20%5Cequiv%20%5Cdfrac%7BA%7D%7Bx%7D%2B%5Cdfrac%7BB%7D%7B%28x-1%29%7D%2B%5Cdfrac%7BC%7D%7B%28x-1%29%5E2%7D%5C%5C%5C%5C%20%5Cimplies%20%5Cdfrac%7Bx%5E2-4%7D%7Bx%28x-1%29%5E2%7D%20%26%20%5Cequiv%20%5Cdfrac%7BA%28x-1%29%5E2%7D%7Bx%28x-1%29%5E2%7D%2B%5Cdfrac%7BBx%28x-1%29%7D%7Bx%28x-1%29%5E2%7D%2B%5Cdfrac%7BCx%7D%7Bx%28x-1%29%5E2%7D%5C%5C%5C%5C%20%5Cimplies%20x%5E2-4%20%26%20%5Cequiv%20A%28x-1%29%5E2%2BBx%28x-1%29%2BCx%20%5Cend%7Baligned%7D)
Calculate the values of A and C using substitution:
![\textsf{when }x=0 \implies -4=A(1)+B(0)+C(0) \implies A=-4](https://tex.z-dn.net/?f=%5Ctextsf%7Bwhen%20%7Dx%3D0%20%5Cimplies%20-4%3DA%281%29%2BB%280%29%2BC%280%29%20%5Cimplies%20A%3D-4)
![\textsf{when }x=1 \implies -3=A(0)+B(0)+C(1) \implies C=-3](https://tex.z-dn.net/?f=%5Ctextsf%7Bwhen%20%7Dx%3D1%20%5Cimplies%20-3%3DA%280%29%2BB%280%29%2BC%281%29%20%5Cimplies%20C%3D-3)
Therefore:
![\begin{aligned}\implies x^2-4 & \equiv -4(x-1)^2+Bx(x-1)-3x\\& \equiv -4(x^2-2x+1)+B(x^2-x)-3x\\& \equiv -4x^2+8x-4+Bx^2-Bx-3x\\& \equiv (B-4)x^2+(5-B)x-4\\\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%20x%5E2-4%20%26%20%5Cequiv%20-4%28x-1%29%5E2%2BBx%28x-1%29-3x%5C%5C%26%20%5Cequiv%20-4%28x%5E2-2x%2B1%29%2BB%28x%5E2-x%29-3x%5C%5C%26%20%5Cequiv%20-4x%5E2%2B8x-4%2BBx%5E2-Bx-3x%5C%5C%26%20%5Cequiv%20%28B-4%29x%5E2%2B%285-B%29x-4%5C%5C%5Cend%7Baligned%7D)
Compare constants to find B:
![\implies 1=B-4 \implies B=5](https://tex.z-dn.net/?f=%5Cimplies%201%3DB-4%20%5Cimplies%20B%3D5)
Substitute the found values of A, B and C:
![\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-\dfrac{3}{(x-1)^2}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bx%5E2-4%7D%7Bx%5E3-2x%5E2%2Bx%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%3D%5Cdisplaystyle%20%5Cint%20-%5Cdfrac%7B4%7D%7Bx%7D%2B%5Cdfrac%7B5%7D%7B%28x-1%29%7D-%5Cdfrac%7B3%7D%7B%28x-1%29%5E2%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
![\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20%5Cdfrac%7B1%7D%7Ba%5En%7D%3Da%5E%7B-n%7D)
![\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-3(x-1)^{-2}}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bx%5E2-4%7D%7Bx%5E3-2x%5E2%2Bx%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%3D%5Cdisplaystyle%20%5Cint%20-%5Cdfrac%7B4%7D%7Bx%7D%2B%5Cdfrac%7B5%7D%7B%28x-1%29%7D-3%28x-1%29%5E%7B-2%7D%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
![\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x$\end{minipage}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cbegin%7Bminipage%7D%7B5%20cm%7D%5Cunderline%7BTerms%20multiplied%20by%20constants%7D%5C%5C%5C%5C%24%5Cdisplaystyle%20%5Cint%20ax%5En%5C%3A%5Ctext%7Bd%7Dx%3Da%20%5Cint%20x%5En%20%5C%3A%5Ctext%7Bd%7Dx%24%5Cend%7Bminipage%7D%7D)
![\boxed{\begin{minipage}{4 cm}\underline{Integrating} $\dfrac{1}{x}$\\\\$\displaystyle \int \dfrac{1}{x}\:\text{d}x=\ln |x|+\text{C}$\end{minipage}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cbegin%7Bminipage%7D%7B4%20cm%7D%5Cunderline%7BIntegrating%7D%20%24%5Cdfrac%7B1%7D%7Bx%7D%24%5C%5C%5C%5C%24%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B1%7D%7Bx%7D%5C%3A%5Ctext%7Bd%7Dx%3D%5Cln%20%7Cx%7C%2B%5Ctext%7BC%7D%24%5Cend%7Bminipage%7D%7D)
![\boxed{\begin{minipage}{4 cm}\underline{Integrating $ax^n$}\\\\$\displaystyle \int ax^n\:\text{d}x=\dfrac{ax^{n+1}}{n+1}+\text{C}$\end{minipage}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cbegin%7Bminipage%7D%7B4%20cm%7D%5Cunderline%7BIntegrating%20%24ax%5En%24%7D%5C%5C%5C%5C%24%5Cdisplaystyle%20%5Cint%20ax%5En%5C%3A%5Ctext%7Bd%7Dx%3D%5Cdfrac%7Bax%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%2B%5Ctext%7BC%7D%24%5Cend%7Bminipage%7D%7D)
![\begin{aligned}\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x & =\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-3(x-1)^{-2}}\:\:\text{d}x\\\\& = \displaystyle -4\int \dfrac{1}{x}\:\:\text{d}x+5\int \dfrac{1}{(x-1)}\:\:\text{d}x-3 \int (x-1)^{-2}}\:\:\text{d}x\\\\& = \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int (x-1)^{-2}}\:\:\text{d}x\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bx%5E2-4%7D%7Bx%5E3-2x%5E2%2Bx%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%3D%5Cdisplaystyle%20%5Cint%20-%5Cdfrac%7B4%7D%7Bx%7D%2B%5Cdfrac%7B5%7D%7B%28x-1%29%7D-3%28x-1%29%5E%7B-2%7D%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%5C%5C%5C%5C%26%20%3D%20%5Cdisplaystyle%20-4%5Cint%20%5Cdfrac%7B1%7D%7Bx%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%2B5%5Cint%20%5Cdfrac%7B1%7D%7B%28x-1%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx-3%20%5Cint%20%28x-1%29%5E%7B-2%7D%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%5C%5C%5C%5C%26%20%3D%20%5Cdisplaystyle%20-4%20%5Cln%20%7Cx%7C%2B5%20%5Cln%7Cx-1%7C-3%20%5Cint%20%28x-1%29%5E%7B-2%7D%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%5Cend%7Baligned%7D)
Use <u>Integration by Substitution</u>:
![\textsf{Let }u=(x-1) \implies \dfrac{\text{d}u}{\text{d}x}=1 \implies \text{d}x=\text{d}u}](https://tex.z-dn.net/?f=%5Ctextsf%7BLet%20%7Du%3D%28x-1%29%20%5Cimplies%20%5Cdfrac%7B%5Ctext%7Bd%7Du%7D%7B%5Ctext%7Bd%7Dx%7D%3D1%20%5Cimplies%20%5Ctext%7Bd%7Dx%3D%5Ctext%7Bd%7Du%7D)
Therefore:
![\implies \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int (x-1)^{-2}}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdisplaystyle%20-4%20%5Cln%20%7Cx%7C%2B5%20%5Cln%7Cx-1%7C-3%20%5Cint%20%28x-1%29%5E%7B-2%7D%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
![\implies \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int u^{-2}}\:\:\text{d}u](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdisplaystyle%20-4%20%5Cln%20%7Cx%7C%2B5%20%5Cln%7Cx-1%7C-3%20%5Cint%20u%5E%7B-2%7D%7D%5C%3A%5C%3A%5Ctext%7Bd%7Du)
![\implies -4 \ln |x|+5 \ln|x-1|-\dfrac{3}{-1}u^{-2+1}+\text{C}](https://tex.z-dn.net/?f=%5Cimplies%20-4%20%5Cln%20%7Cx%7C%2B5%20%5Cln%7Cx-1%7C-%5Cdfrac%7B3%7D%7B-1%7Du%5E%7B-2%2B1%7D%2B%5Ctext%7BC%7D)
![\implies -4 \ln |x|+5 \ln|x-1|+3u^{-1}+\text{C}](https://tex.z-dn.net/?f=%5Cimplies%20-4%20%5Cln%20%7Cx%7C%2B5%20%5Cln%7Cx-1%7C%2B3u%5E%7B-1%7D%2B%5Ctext%7BC%7D)
![\implies -4 \ln |x|+5 \ln|x-1|+\dfrac{3}{u}+\text{C}](https://tex.z-dn.net/?f=%5Cimplies%20-4%20%5Cln%20%7Cx%7C%2B5%20%5Cln%7Cx-1%7C%2B%5Cdfrac%7B3%7D%7Bu%7D%2B%5Ctext%7BC%7D)
Substitute back in u = (x - 1):
![\implies -4 \ln |x|+5 \ln|x-1|+\dfrac{3}{x-1}+\text{C}](https://tex.z-dn.net/?f=%5Cimplies%20-4%20%5Cln%20%7Cx%7C%2B5%20%5Cln%7Cx-1%7C%2B%5Cdfrac%7B3%7D%7Bx-1%7D%2B%5Ctext%7BC%7D)
Learn more about integration here:
brainly.com/question/27988986
brainly.com/question/27805589