Answer: 55 Square inches
Step-by-step explanation:
You should find the area of the square and rectangle separately
To find the area you multiply length times width, so the area of the square is 4×4 which equals 16
Then you do the same to the rectangle, 13×3=39
To find the area of the whole figure you add the area of the two shapes together 16+39 which should give you 55
Answer:
![A=256\ units^2](https://tex.z-dn.net/?f=A%3D256%5C%20units%5E2)
Step-by-step explanation:
The given figure shows a triangular pyramid having 16 units as height. The base of pyramid is triangle.
The area of a triangular pyramid is given by :
![A=\dfrac{1}{3}bH](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B1%7D%7B3%7DbH)
Where
b is the area of base
![A=\dfrac{1}{3}\times \dfrac{1}{2}\times B\times h\times H\\\\=\dfrac{1}{6}\times 12\times 8\times 16\\\\=256\ units^2](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20B%5Ctimes%20h%5Ctimes%20H%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B6%7D%5Ctimes%2012%5Ctimes%208%5Ctimes%2016%5C%5C%5C%5C%3D256%5C%20units%5E2)
So, the required area is equal to
.
Answer:
Option D (r(t) = 3.50t +25
; r(8) = 53)
Step-by-step explanation:
The fixed cost to rent the kayak $25. This is the cost which remains fixed irrespective of the usage of the kayak. The variable cost of using the kayak is the cost which depends on the usage of the kayak. It is mentioned that the kayak is used for 4 hours and the company charges $3.5 for every half hour. The cost function is given by:
r(t) = 25 + 3.5t ; there r is the total cost of using the kayak and t is the number of half-hours the kayak is used.
4 hours means that there are 8 half-hours. Therefore, t=8. Put t=8 in r(t).
r(8) = 25 + 3.5*(8) = 25 + 28 = 53.
Therefore, Option D is the correct answer!!!
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)