9514 1404 393
Answer:
-7x^2 +8x -10
Step-by-step explanation:
Combine like terms in the usual way.
(-3x^2 +5x -8) -(4x^2 -3x +2)
= -3^2 +5x -8 -4x^2 +3x -2
= (-3-4)x^2 +(5+3)x +(-8-2)
= -7x^2 +8x -10
Answer:
blue question mark = –
Step-by-step explanation:
6(2x -1) uses the distributive property, which means that both 2x and -1 are multiplied by 6 when expanding the brackets.
6(2x - 1)
= (6 x 2x) + (6 x -1)
= 12x - 6
Answer:
The rate at which the temperature is changing at 4 p.m. is ![T'(16)=-\frac{5\pi }{12}\approx-1.309 \:\frac{\ºF}{h}](https://tex.z-dn.net/?f=T%27%2816%29%3D-%5Cfrac%7B5%5Cpi%20%7D%7B12%7D%5Capprox-1.309%20%5C%3A%5Cfrac%7B%5C%C2%BAF%7D%7Bh%7D)
Step-by-step explanation:
The derivative,
is the instantaneous rate of change of
with respect to
when
.
We know that the daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function
,
where
is hours after midnight.
To find the rate at which the temperature is changing at 4 p.m. First, we must find the derivative of this function.
![T'(x)=\frac{d}{dx}\left(94-10\cos \left[\frac{\pi \:\:}{12}\left(x-2\:\right)\right]\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g\\\\T'(x)=\frac{d}{dx}\left(94\right)-\frac{d}{dx}\left(10\cos \left(\frac{\pi }{12}\left(x-2\right)\right)\right)](https://tex.z-dn.net/?f=T%27%28x%29%3D%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%2894-10%5Ccos%20%5Cleft%5B%5Cfrac%7B%5Cpi%20%5C%3A%5C%3A%7D%7B12%7D%5Cleft%28x-2%5C%3A%5Cright%29%5Cright%5D%5Cright%29%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Athe%5C%3ASum%2FDifference%5C%3ARule%7D%3A%5Cquad%20%5Cleft%28f%5Cpm%20g%5Cright%29%27%3Df%5C%3A%27%5Cpm%20g%5C%5C%5C%5CT%27%28x%29%3D%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%2894%5Cright%29-%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%2810%5Ccos%20%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%28x-2%5Cright%29%5Cright%29%5Cright%29)
![\mathrm{Derivative\:of\:a\:constant}:\quad \frac{d}{dx}\left(a\right)=0\\\\\frac{d}{dx}\left(94\right)=0](https://tex.z-dn.net/?f=%5Cmathrm%7BDerivative%5C%3Aof%5C%3Aa%5C%3Aconstant%7D%3A%5Cquad%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%28a%5Cright%29%3D0%5C%5C%5C%5C%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%2894%5Cright%29%3D0)
![\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\f=\cos \left(u\right),\:\:u=\frac{\pi }{12}\left(x-2\right)\\\\](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Athe%5C%3Achain%5C%3Arule%7D%3A%5Cquad%20%5Cfrac%7Bdf%5Cleft%28u%5Cright%29%7D%7Bdx%7D%3D%5Cfrac%7Bdf%7D%7Bdu%7D%5Ccdot%20%5Cfrac%7Bdu%7D%7Bdx%7D%5C%5Cf%3D%5Ccos%20%5Cleft%28u%5Cright%29%2C%5C%3A%5C%3Au%3D%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%28x-2%5Cright%29%5C%5C%5C%5C)
![10\frac{d}{du}\left(\cos \left(u\right)\right)\frac{d}{dx}\left(\frac{\pi }{12}\left(x-2\right)\right)\\\\10\left(-\sin \left(u\right)\right)\frac{\pi }{12}\\\\\mathrm{Substitute\:back}\:u=\frac{\pi }{12}\left(x-2\right)\\\\10\left(-\sin \left(\frac{\pi }{12}\left(x-2\right)\right)\right)\frac{\pi }{12}\\\\\frac{d}{dx}\left(10\cos \left(\frac{\pi }{12}\left(x-2\right)\right)\right)=-\frac{5\pi }{6}\sin \left(\frac{\pi }{12}\left(x-2\right)\right)](https://tex.z-dn.net/?f=10%5Cfrac%7Bd%7D%7Bdu%7D%5Cleft%28%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%29%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%28x-2%5Cright%29%5Cright%29%5C%5C%5C%5C10%5Cleft%28-%5Csin%20%5Cleft%28u%5Cright%29%5Cright%29%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5C%5C%5C%5C%5Cmathrm%7BSubstitute%5C%3Aback%7D%5C%3Au%3D%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%28x-2%5Cright%29%5C%5C%5C%5C10%5Cleft%28-%5Csin%20%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%28x-2%5Cright%29%5Cright%29%5Cright%29%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5C%5C%5C%5C%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%2810%5Ccos%20%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%28x-2%5Cright%29%5Cright%29%5Cright%29%3D-%5Cfrac%7B5%5Cpi%20%7D%7B6%7D%5Csin%20%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%28x-2%5Cright%29%5Cright%29)
So,
![T'(x)=\frac{5\pi }{6}\sin \left(\frac{\pi }{12}\left(x-2\right)\right)](https://tex.z-dn.net/?f=T%27%28x%29%3D%5Cfrac%7B5%5Cpi%20%7D%7B6%7D%5Csin%20%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%28x-2%5Cright%29%5Cright%29)
Then, at 4 p.m means
because
is hours after midnight.
Thus,
![T'(16)=\frac{5\pi }{6}\sin \left(\frac{\pi }{12}\left(16-2\right)\right)\\\\T'(16)=-\frac{5\pi }{12}\approx-1.309](https://tex.z-dn.net/?f=T%27%2816%29%3D%5Cfrac%7B5%5Cpi%20%7D%7B6%7D%5Csin%20%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B12%7D%5Cleft%2816-2%5Cright%29%5Cright%29%5C%5C%5C%5CT%27%2816%29%3D-%5Cfrac%7B5%5Cpi%20%7D%7B12%7D%5Capprox-1.309)
The rate at which the temperature is changing at 4 p.m. is ![T'(16)=-\frac{5\pi }{12}\approx-1.309 \:\frac{\ºF}{h}](https://tex.z-dn.net/?f=T%27%2816%29%3D-%5Cfrac%7B5%5Cpi%20%7D%7B12%7D%5Capprox-1.309%20%5C%3A%5Cfrac%7B%5C%C2%BAF%7D%7Bh%7D)
Answer:
sorry but need more info y could be any number