Answer:
a₄ = 12
Step-by-step explanation:
To obtain the terms in the sequence add 2 to the previous term, that is
a₂ = a₁ + 2 = 6 + 2 = 8
a₃ = a₂ + 2 = 8 + 2 = 10
a₄ = a₃ + 2 = 10 + 2 = 12
Answer:
3(3 +4n)
OR
9 + 12n
Step-by-step explanation:
9 + 12n
Factorize it
3(3 +4n)
OR
The answer is the same.
Since we can't add a number with a number and letter. So in that case the answer is ;
9 + 12n.
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Parameterize the line segments (call them
and
, respectively, by
![\vec r_1(t)=(1-t)(0,0)+t(4,1)=(4t,t)](https://tex.z-dn.net/?f=%5Cvec%20r_1%28t%29%3D%281-t%29%280%2C0%29%2Bt%284%2C1%29%3D%284t%2Ct%29)
![\vec r_2(t)=(1-t)(4,1)+t(5,0)=(4+t,1-t)](https://tex.z-dn.net/?f=%5Cvec%20r_2%28t%29%3D%281-t%29%284%2C1%29%2Bt%285%2C0%29%3D%284%2Bt%2C1-t%29)
both with
. Then
![\displaystyle\int_C(x+4y)\,\mathrm dx+x^2\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%28x%2B4y%29%5C%2C%5Cmathrm%20dx%2Bx%5E2%5C%2C%5Cmathrm%20dy)
![=\displaystyle\int_0^1\bigg((4t+4t)(4)+(4t)^2(1)\bigg)\,\mathrm dt+\int_0^1\bigg((4+t+4(1-t))(1)+(4+t)^2(-1)\bigg)\,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E1%5Cbigg%28%284t%2B4t%29%284%29%2B%284t%29%5E2%281%29%5Cbigg%29%5C%2C%5Cmathrm%20dt%2B%5Cint_0%5E1%5Cbigg%28%284%2Bt%2B4%281-t%29%29%281%29%2B%284%2Bt%29%5E2%28-1%29%5Cbigg%29%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^115t^2+21t-8\,\mathrm dt=\boxed{\frac{15}2}](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E115t%5E2%2B21t-8%5C%2C%5Cmathrm%20dt%3D%5Cboxed%7B%5Cfrac%7B15%7D2%7D)
I will assume that you meant:
(x+4)/(-3x^2+12x+36) factor the denominator...
(x+4)/(-3(x^2-4x-12))
To factor a quadratic of the form ax^2+bx+c you need to find two values, j and k, which satisfy two conditions...
jk=ac=-12 and j+k=b=-4 so j and k must be 2 and -6 and the factors are then:
(x+2)(x-6) so we are now left with:
(x+4)/(-3(x+2)(x-6))
The only restriction on this function is that division by zero is undefined, so x cannot equal -2 or 6