Answer:
I think it's c tell me it i'm wrong
Step-by-step explanation:
they both have and so it would be times
Answer:
The distance would be about 32 (31.62) feet.
Step-by-step explanation:
∛0.008 = ∛(0.2)^3
= 0.2
hope it helps
The trapezoidal approximation will be the average of the left- and right-endpoint approximations.
Let's consider a simple example of estimating the value of a general definite integral,
![\displaystyle\int_a^bf(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_a%5Ebf%28x%29%5C%2C%5Cmathrm%20dx)
Split up the interval
![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
into
![n](https://tex.z-dn.net/?f=n)
equal subintervals,
![[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]](https://tex.z-dn.net/?f=%5Bx_0%2Cx_1%5D%5Ccup%5Bx_1%2Cx_2%5D%5Ccup%5Ccdots%5Ccup%5Bx_%7Bn-2%7D%2Cx_%7Bn-1%7D%5D%5Ccup%5Bx_%7Bn-1%7D%2Cx_n%5D)
where
![a=x_0](https://tex.z-dn.net/?f=a%3Dx_0)
and
![b=x_n](https://tex.z-dn.net/?f=b%3Dx_n)
. Each subinterval has measure (width)
![\dfrac{a-b}n](https://tex.z-dn.net/?f=%5Cdfrac%7Ba-b%7Dn)
.
Now denote the left- and right-endpoint approximations by
![L](https://tex.z-dn.net/?f=L)
and
![R](https://tex.z-dn.net/?f=R)
, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are
![\{x_0,x_1,\cdots,x_{n-1}\}](https://tex.z-dn.net/?f=%5C%7Bx_0%2Cx_1%2C%5Ccdots%2Cx_%7Bn-1%7D%5C%7D)
. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints,
![\{x_1,x_2,\cdots,x_n\}](https://tex.z-dn.net/?f=%5C%7Bx_1%2Cx_2%2C%5Ccdots%2Cx_n%5C%7D)
.
So, you have
![L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)](https://tex.z-dn.net/?f=L%3D%5Cdfrac%7Bb-a%7Dn%5Cleft%28f%28x_0%29%2Bf%28x_1%29%2B%5Ccdots%2Bf%28x_%7Bn-2%7D%29%2Bf%28x_%7Bn-1%7D%29%5Cright%29)
![R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7Bb-a%7Dn%5Cleft%28f%28x_1%29%2Bf%28x_2%29%2B%5Ccdots%2Bf%28x_%7Bn-1%7D%29%2Bf%28x_n%29%5Cright%29)
Now let
![T](https://tex.z-dn.net/?f=T)
denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,
![T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7Bb-a%7Dn%5Cleft%28%5Cdfrac%7Bf%28x_0%29%2Bf%28x_1%29%7D2%2B%5Cdfrac%7Bf%28x_1%29%2Bf%28x_2%29%7D2%2B%5Ccdots%2B%5Cdfrac%7Bf%28x_%7Bn-2%7D%29%2Bf%28x_%7Bn-1%7D%29%7D2%2B%5Cdfrac%7Bf%28x_%7Bn-1%7D%29%2Bf%28x_n%29%7D2%5Cright%29)
Factoring out
![\dfrac12](https://tex.z-dn.net/?f=%5Cdfrac12)
and regrouping the terms, you have
![T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7Bb-a%7D%7B2n%7D%5Cleft%28%28f%28x_0%29%2Bf%28x_1%29%2B%5Ccdots%2Bf%28x_%7Bn-2%7D%29%2Bf%28x_%7Bn-1%7D%29%29%2B%28f%28x_1%29%2Bf%28x_2%29%2B%5Ccdots%2Bf%28x_%7Bn-1%7D%29%2Bf%28x_n%29%29%5Cright%29)
which is equivalent to
![T=\dfrac12\left(L+R)](https://tex.z-dn.net/?f=T%3D%5Cdfrac12%5Cleft%28L%2BR%29)
and is the average of
![L](https://tex.z-dn.net/?f=L)
and
![R](https://tex.z-dn.net/?f=R)
.
So the trapezoidal approximation for your problem should be
Answer:
(1, 3)
Step-by-step explanation:
x - 3y = -8
3x + y = 6
Isolate a variable in one of the equations:
y = 6 - 3x
Substitute the value of y into the other equation:
x - 3(6 - 3x) = -8
Use distributive property:
x - 18 + 9x = -8
Combine like terms:
10x - 18 = -8
Isolate the variable:
10x = 10
x = 1
Substitute the value of x into any equation:
3(1) + y = 6
3 + y = 6
Isolate the variable:
y = 3