Answer:
y= 1200(0.92)^x
Step-by-step explanation:
Given data
Cost price= $1200
Rate of decrease= 8%
Let the time be x
The expression that describes an exponential function is
y= ab^x
for decrease b= 1-r
y= 1200(1-0.08)^x
y= 1200(0.92)^x
Hence the expression is
y= 1200(0.92)^x
Answer:
The other coordinate is (-6,19)
Step-by-step explanation:
![m = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2} )](https://tex.z-dn.net/?f=m%20%3D%20%28%20%5Cfrac%7Bx1%20%2B%20x2%7D%7B2%7D%20%2C%20%5Cfrac%7By1%20%2B%20y2%7D%7B2%7D%20%29)
(4,-7)
(x,y)
[(4+x)/2 , (-7+y)/2] = (-1,6)
(4 + x)/2 = -1
4 + x = -2
× = -6
(-7 + y)/2 = 6
-7 + y = 12
y = 19
(Correct me if i am wrong)
Answer:
2758 Nm
Step-by-step explanation
Work done usually depends on two things force applied and distance travelled due to applied force. In this current scenario, the force is being applied at an angle so we will have to find a component of force in the direction of the movement.
We usually find component using cos θ.
Here θ is 40°
Now, the modified equation becomes,
Work Done = Force * Distance * Component of force along the direction of distance
∴ Work = 30 N * 120 m * Cos 40°
⇒Work = 30 * 120 * 0.766
⇒Work = 2757.6 Nm
Rounding to the nearest whole number.
∴ The work done by force is 2758 Nm which is option B
Answer:
<h2>
![\bold{a_7=33x-15}](https://tex.z-dn.net/?f=%5Cbold%7Ba_7%3D33x-15%7D)
</h2>
Step-by-step explanation:
In arithmetic sequence: a₂ - a₁ = a₃ - a₂ {common difference}
![a_1=5x+9\\a_2=8x+5\\a_3=11x+1\\\\\\d=8x+5-(5x+9)=8x+5-5x-9=3x-4\\\\a_n=a_1+d(n-1)\\\\a_7=5x+9+(3x-4)\cdot6\\\\a_7=5x+9+18x-24\\\\a_7=33x-15](https://tex.z-dn.net/?f=a_1%3D5x%2B9%5C%5Ca_2%3D8x%2B5%5C%5Ca_3%3D11x%2B1%5C%5C%5C%5C%5C%5Cd%3D8x%2B5-%285x%2B9%29%3D8x%2B5-5x-9%3D3x-4%5C%5C%5C%5Ca_n%3Da_1%2Bd%28n-1%29%5C%5C%5C%5Ca_7%3D5x%2B9%2B%283x-4%29%5Ccdot6%5C%5C%5C%5Ca_7%3D5x%2B9%2B18x-24%5C%5C%5C%5Ca_7%3D33x-15)
I'm reading this as
![\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C2xe%5E%7B-y%7D%5C%2C%5Cmathrm%20dx%2B%282y-x%5E2e%5E%7B-y%7D%29%5C%2C%5Cmathrm%20dy)
with
![\nabla f=(2xe^{-y},2y-x^2e^{-y})](https://tex.z-dn.net/?f=%5Cnabla%20f%3D%282xe%5E%7B-y%7D%2C2y-x%5E2e%5E%7B-y%7D%29)
.
The value of the integral will be independent of the path if we can find a function
![f(x,y)](https://tex.z-dn.net/?f=f%28x%2Cy%29)
that satisfies the gradient equation above.
You have
![\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%3D2xe%5E%7B-y%7D%5C%5C%5C%5C%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3D2y-x%5E2e%5E%7B-y%7D%5Cend%7Bcases%7D)
Integrate
![\dfrac{\partial f}{\partial x}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D)
with respect to
![x](https://tex.z-dn.net/?f=x)
. You get
![\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%5C%2C%5Cmathrm%20dx%3D%5Cint2xe%5E%7B-y%7D%5C%2C%5Cmathrm%20dx)
![f=x^2e^{-y}+g(y)](https://tex.z-dn.net/?f=f%3Dx%5E2e%5E%7B-y%7D%2Bg%28y%29)
Differentiate with respect to
![y](https://tex.z-dn.net/?f=y)
. You get
![\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3D%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5Bx%5E2e%5E%7B-y%7D%2Bg%28y%29%5D)
![2y-x^2e^{-y}=-x^2e^{-y}+g'(y)](https://tex.z-dn.net/?f=2y-x%5E2e%5E%7B-y%7D%3D-x%5E2e%5E%7B-y%7D%2Bg%27%28y%29)
![2y=g'(y)](https://tex.z-dn.net/?f=2y%3Dg%27%28y%29)
Integrate both sides with respect to
![y](https://tex.z-dn.net/?f=y)
to arrive at
![\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint2y%5C%2C%5Cmathrm%20dy%3D%5Cint%20g%27%28y%29%5C%2C%5Cmathrm%20dy)
![y^2=g(y)+C](https://tex.z-dn.net/?f=y%5E2%3Dg%28y%29%2BC)
![g(y)=y^2+C](https://tex.z-dn.net/?f=g%28y%29%3Dy%5E2%2BC)
So you have
![f(x,y)=x^2e^{-y}+y^2+C](https://tex.z-dn.net/?f=f%28x%2Cy%29%3Dx%5E2e%5E%7B-y%7D%2By%5E2%2BC)
The gradient is continuous for all
![x,y](https://tex.z-dn.net/?f=x%2Cy)
, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is