Answer:
BA = 25π,
LA = 25√2π,
TA = 25π + 25√2π,
V = 41 and 2 / 3π
Step-by-step explanation:
We need to determine the height here, as it is not given, and is quite important to us. The height is a perpendicular line segment to the radius, hence forming a 45 - 45 - 90 degree triangle as you can see. Therefore, by " Converse to Base Angles Theorem " the height should be equal in length to the radius,
( Height = 5 inches = Radius
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Now knowing the height, let's begin by calculating the base area. By it's name, we have to find the area of the base. As it is a circle, let us apply the formula " πr^2 "
- Base Area = 25π
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The lateral area is simply the surface area excluding the base area, the surface area having a formula of " πr^2 + πrl. " Thus, the lateral area can be calculated through the formula " πrl, " but as we are not given the slant height ( l ) we have to use another formula,
-
- Lateral Area = 25√2π
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And the surface area is the base area + lateral area -
- Surface Area
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The volume of a cone is 1 / 3rd that of a cylinder, with a simple formula of Base * height. Therefore, we can conclude the following -
- Volume = 41 and 2 / 3π
Answer:
Slope intercept: y = -3/2x + 1
Point slope: y + 2 = -3/2 * (x - 2) [Forgot to add the work for this, I will add it if you need it, feel free to ask.]
Step-by-step explanation:
m = (change in y)/change in x)
But also
m = y_2 - y_1/x_2 - x_1
So lets substitute
m = 1 - (-2)/0 - (2)
Lets find the slope
m = 3/0 - (2)
m = 3/-2
m = -3/2 (Moved the negative)
Now we find the value of b using the equation of a line.
y = mx + b
y = (-3/2) * x + b
y = (-3/2) * (2) + b
-2 = (-3/2) * (2) + b
Now we find the value of b
Lets rewrite
-3/2 * 2 + b = -2
Cancel the CF of 2
-3 + b = -2
Move the terms without b to the right
b = -2 + 3
b = 1
Now we substitute our values of the slope and y-int into y = mx + b to find the equation.
y = -3/2x + 1
The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is
![\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_C%5Cmathrm%20ds%20%3D%20%5Cint_a%5Eb%20%5Csqrt%7B%5Cleft%28%5Cfrac%7B%5Cmathrm%20dx%7D%7B%5Cmathrm%20dt%7D%5Cright%29%5E2%2B%5Cleft%28%5Cfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D%5Cright%29%5E2%7D%20%5C%2C%5Cmathrm%20dt)
In this case, we have
<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> ) ==> d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )
<em>y(t)</em> = 5 - 2<em>t</em> ==> d<em>y</em>/d<em>t</em> = -2
and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then
![\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E2%20%5Csqrt%7B%5Cleft%28e%5Et-e%5E%7B-t%7D%5Cright%29%5E2%2B%28-2%29%5E2%7D%20%5C%2C%5Cmathrm%20dt%20%3D%20%5Cint_0%5E2%20%5Csqrt%7Be%5E%7B2t%7D-2%2Be%5E%7B-2t%7D%2B4%7D%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%20%5Csqrt%7Be%5E%7B2t%7D%2B2%2Be%5E%7B-2t%7D%7D%20%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Csqrt%7B%5Cleft%28e%5Et%2Be%5E%7B-t%7D%5Cright%29%5E2%7D%20%5C%2C%5Cmathrm%20dt)
![=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Cleft%28e%5Et%2Be%5E%7B-t%7D%5Cright%29%5C%2C%5Cmathrm%20dt)
![=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}](https://tex.z-dn.net/?f=%3D%5Cleft%28e%5Et-e%5E%7B-t%7D%5Cright%29%5Cbigg%7C_0%5E2%20%3D%20%5Cleft%28e%5E2-e%5E%7B-2%7D%5Cright%29-%5Cleft%28e%5E0-e%5E%7B-0%7D%5Cright%29%20%3D%20%5Cboxed%7Be%5E2-%5Cfrac1%7Be%5E2%7D%7D)
Yes it is a square number