Answer:
1480cm³
Step-by-step explanation:
The Equation for the Volume of a Cylinder is V=Bh, where B is the area of the circle (A= πr²).
We are given all of the things we need, we just have to plug them into the equation.
First let's find B using the radius(r) which is given to be 6.4cm, so B=πr²
B=π(6.4cm)²
Plug that into a calculator to get 128.6796351cm²
Now we can plug that into the original equation, V=Bh.
V=(128.6796351cm²)h
h is also given to be 11.5cm, so lets plug that in.
V=(128.6796351cm²)(11.5cm)
You will get V=1479.815804cm³, but since we are rounding to the nearest cubic centimeter, we would round up to 1480cm³.
Answer:
The answer is below
Step-by-step explanation:
We need to prove that:
(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.
Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.
Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A
![\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7BsecA-1%7D%20%7D%7B%5Csqrt%7BsecA%2B1%7D%20%7D%20%2B%5Cfrac%7B%5Csqrt%7BsecA%2B1%7D%20%7D%7B%5Csqrt%7BsecA-1%7D%20%7D%20%3D%5Cfrac%7B%28%5Csqrt%7BsecA-1%29%7D%28%5Csqrt%7BsecA-1%7D%29%2B%28%5Csqrt%7BsecA%2B1%29%7D%28%5Csqrt%7BsecA%2B1%7D%29%20%7D%7B%28%5Csqrt%7BsecA%2B1%7D%29%28%5Csqrt%7BsecA-1%7D%29%20%7D%20%5C%5C%5C%5C%3D%5Cfrac%7BsecA-1%2B%28secA%2B1%29%7D%7B%5Csqrt%7Bsec%5E2A-secA%2BsecA-1%7D%20%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2secA%7D%7B%5Csqrt%7Bsec%5E2A-1%7D%20%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2secA%7D%7B%5Csqrt%7Btan%5E2A%7D%20%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2secA%7D%7BtanA%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B2%2A%5Cfrac%7B1%7D%7BcosA%7D%20%7D%7B%5Cfrac%7BsinA%7D%7BcosA%7D%20%7D%5C%5C%5C%5C%3D%202%2A%5Cfrac%7B1%7D%7BcosA%7D%2A%5Cfrac%7BcosA%7D%7BsinA%7D%5C%5C%5C%5C%3D2%2A%5Cfrac%7B1%7D%7BsinAA%7D%5C%5C%5C%5C%3D2cosecA)
1/3 is the slope, or 1 over 3.