Answer:
x=-8 or x=-1 (first option)
Step-by-step explanation:
![\sqrt{1-3x}=x+3\\1-3x=(x+3)^{2} \\](https://tex.z-dn.net/?f=%5Csqrt%7B1-3x%7D%3Dx%2B3%5C%5C1-3x%3D%28x%2B3%29%5E%7B2%7D%20%5C%5C)
now expand the bracket on the right
![1-3x=x^{2} +6x+9\\simplify:\\1-9=x^2+6x+3x\\-8=x^2+9x\\](https://tex.z-dn.net/?f=1-3x%3Dx%5E%7B2%7D%20%2B6x%2B9%5C%5Csimplify%3A%5C%5C1-9%3Dx%5E2%2B6x%2B3x%5C%5C-8%3Dx%5E2%2B9x%5C%5C)
now make it a quadratic equation:
![x^2+9x+8=0\\factorise:\\(x+1)(x+8)\\so:\\x+1=0 \\x=-1\\\\and \\x+8=0\\x=-8](https://tex.z-dn.net/?f=x%5E2%2B9x%2B8%3D0%5C%5Cfactorise%3A%5C%5C%28x%2B1%29%28x%2B8%29%5C%5Cso%3A%5C%5Cx%2B1%3D0%20%5C%5Cx%3D-1%5C%5C%5C%5Cand%20%5C%5Cx%2B8%3D0%5C%5Cx%3D-8)
Answer:
Step-by-step explanation:
Hello,
as cosx+sinx=k we can write that
![(cosx+sinx)^2=cos^2x+sin^2x+2\ cosx \ sinx = k^2](https://tex.z-dn.net/?f=%28cosx%2Bsinx%29%5E2%3Dcos%5E2x%2Bsin%5E2x%2B2%5C%20cosx%20%5C%20sinx%20%3D%20k%5E2)
and we know that ![cos^2x+sin^2x=1](https://tex.z-dn.net/?f=cos%5E2x%2Bsin%5E2x%3D1)
so
![2 \ cosx \ sinx \ = k^2 -1 \\ sinx \ cosx = \dfrac{k^2-1}{2}](https://tex.z-dn.net/?f=2%20%5C%20cosx%20%5C%20sinx%20%5C%20%3D%20k%5E2%20-1%20%5C%5C%3C%3D%3E%20sinx%20%5C%20cosx%20%3D%20%5Cdfrac%7Bk%5E2-1%7D%7B2%7D)
which is the answer to the question 2
Now, let s estimate
![1=1^2=(cos^2x+sin^2x)^2=cos^4x+sin^4x+2cos^2xsin^2x](https://tex.z-dn.net/?f=1%3D1%5E2%3D%28cos%5E2x%2Bsin%5E2x%29%5E2%3Dcos%5E4x%2Bsin%5E4x%2B2cos%5E2xsin%5E2x)
so
![cos^4x+sin^4x=1-2cos^2xsin^2x](https://tex.z-dn.net/?f=cos%5E4x%2Bsin%5E4x%3D1-2cos%5E2xsin%5E2x)
We use the previous result to write
![cos^4x+sin^4x=1-2cos^2xsin^2x = 1-2(\dfrac{k^2-1}{2})^2 = \dfrac{2-(k^2-1)^2}{2}](https://tex.z-dn.net/?f=cos%5E4x%2Bsin%5E4x%3D1-2cos%5E2xsin%5E2x%20%3D%201-2%28%5Cdfrac%7Bk%5E2-1%7D%7B2%7D%29%5E2%20%3D%20%5Cdfrac%7B2-%28k%5E2-1%29%5E2%7D%7B2%7D)
and we know that
![(k^2-1)^2=k^4-2k^2+1](https://tex.z-dn.net/?f=%28k%5E2-1%29%5E2%3Dk%5E4-2k%5E2%2B1)
so
![cos^4x+sin^4x=\dfrac{2-k^4+2k^2-1}{2}=\dfrac{-k^4+2k^2+1}{2}](https://tex.z-dn.net/?f=cos%5E4x%2Bsin%5E4x%3D%5Cdfrac%7B2-k%5E4%2B2k%5E2-1%7D%7B2%7D%3D%5Cdfrac%7B-k%5E4%2B2k%5E2%2B1%7D%7B2%7D)
this is the answer to the first question
finally, let s estimate
so ![(sinx-cosx)=\sqrt{2-k^2}](https://tex.z-dn.net/?f=%28sinx-cosx%29%3D%5Csqrt%7B2-k%5E2%7D)
and this is the answer to the last question
do not hesitate if you need further explanation
hope this helps
Answer:
Thx for the free points! Have a great day
Step-by-step explanation:
Answer:
It will take the object 2 seconds to reach the maximum height.
Step-by-step explanation:
The original equation is
.
The object reaches the maximum height when derivative of
is zero
![\frac{dh}{dt}=0](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%3D0)
So let us take evaluate the derivative of ![h](https://tex.z-dn.net/?f=h)
![\frac{dh}{dt} =\frac{d}{dt} (-14t^2+56t+60)=\frac{d}{dt} (-14t^2)+\frac{d}{dt}(56t)+\frac{d}{dt} (60)](https://tex.z-dn.net/?f=%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D%5Cfrac%7Bd%7D%7Bdt%7D%20%28-14t%5E2%2B56t%2B60%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%20%28-14t%5E2%29%2B%5Cfrac%7Bd%7D%7Bdt%7D%2856t%29%2B%5Cfrac%7Bd%7D%7Bdt%7D%20%2860%29)
![=-28t+56](https://tex.z-dn.net/?f=%3D-28t%2B56)
this must be equal to 0:
![-28t+56=0](https://tex.z-dn.net/?f=-28t%2B56%3D0)
![t=\frac{-56}{-28}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-56%7D%7B-28%7D)
![\boxed{ t=2}](https://tex.z-dn.net/?f=%5Cboxed%7B%20t%3D2%7D)
It will take the object 2 seconds to reach the maximum height.