Answer: $4.40
Step-by-step explanation:
Trust me did the assignment
Answer:
Step-by-step explanation:
The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.
For the integral from [0, 2], the equation of the line is -3x + 6;
For the integral from [2, 3], the equation of the line is 3x - 6.
We integrate then:
and
sorry for the odd representation; that's as good as it gets here!
Using the First Fundamental Theorem of Calculus, we get:
(6 - 0) + (-4.5 - (-6)) = 6 + 1.5 = 7.5
So in matt's equation, he made a mistake in the a transision from line 2 to line 3
in line 2: -4(-2)2
in line 3: -4(4)
the mistake is that -2 times 2 is not equal +4 it is equal to -4
also from lines 5 to 6 he made a mistake in order of opperations (mulit division then addition and subtract)
line 5: -10+30/5
line 6: 20/5
so he first subtracted 10 then divided, he should have divided then subtracted
so the equation should have equaled
Karen used the correct (-) times (+) property and the order of operations
so Karen is correct and Matt is wrong.
Ok so to find which sides are congruent we need to know their lengths.
To find the length we need the distance formula between two point ->
√(X2-X1)∧2 +(Y2-Y1)∧2
Ok lets find the first side PQ
P(-1,3) Q(2,-1)
X1 Y1 X2 Y2
√(2-(-1)∧2 + (-1-3)∧2 = 5
Now PR
P (-1,3) R (5,3)
X1 Y1 X2 Y2
√(5-(-1))∧2 + (3-3)∧2) = 6
Now the last side QR
Q (2, -1) R (5,3)
X1 Y1 X2 Y2
√(5-2)∧2 + (3-(-1))∧2 = 5
From the above work we see that PQ and QR are congruent becuase they are equal PQ=QR
Also the opposite angles of these sides are congruent. Hope this helps :).
first off let's notice that the height is 11 meters and the volume of the cone is 103.62 cubic centimeters, so let's first convert the height to the corresponding unit for the volume, well 1 meters is 100 cm, so 11 m is 1100 cm.
![\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=\stackrel{cm^3}{103.62}\\ h=\stackrel{cm}{1100} \end{cases}\implies 103.62=\cfrac{\pi r^2 (1100)}{3} \\\\\\ 3(103.62)=1100\pi r^2\implies \cfrac{3(103.62)}{1100\pi }=r^2 \\\\\\ \sqrt{\cfrac{3(103.62)}{1100\pi }}=r\implies \stackrel{cm}{0.00510199305952} \approx r](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20V%3D%5Cstackrel%7Bcm%5E3%7D%7B103.62%7D%5C%5C%20h%3D%5Cstackrel%7Bcm%7D%7B1100%7D%20%5Cend%7Bcases%7D%5Cimplies%20103.62%3D%5Ccfrac%7B%5Cpi%20r%5E2%20%281100%29%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%203%28103.62%29%3D1100%5Cpi%20r%5E2%5Cimplies%20%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%3Dr%5E2%20%5C%5C%5C%5C%5C%5C%20%5Csqrt%7B%5Ccfrac%7B3%28103.62%29%7D%7B1100%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Cstackrel%7Bcm%7D%7B0.00510199305952%7D%20%5Capprox%20r)