Answer:
the last one: yes it can, because -4.5 lies to left of -3.5
Step-by-step explanation:
(1) <span>The surface area of the square pyramid = 4 * (area of one face) + area of the base
area of one face = 0.5 * 34.2 * 28.4
area of the base = 34.2 * 34.2
∴ </span>The surface area of the pyramid = 4 * (<span>0.5 * 34.2 * 28.4) + </span><span>34.2 * 34.2
By comparing the last answer with the answer of </span>Vikram, we find that:
He used the wrong expression to represent the area of the base of the pyramid.
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(2) The surface area of the rectangular pyramid
= area of two face with the height 36.5 + area of two face with the height 37.8 + area of the base
area of two face with the height 36.5 = 2 * 0.5 * 36.5 * 25.6
area of two face with the height 37.8 = 2 * 0.5 * 37.8 * 16.2
are of the base = 25.6 * 16.2
Total area = (2 * 0.5 * 36.5 * 25.6) + (2 * 0.5 * 37.8 * 16.2) + (25.6 * 16.2)
= (36.5 * 25.6) + (37.8 * 16.2) + (25.6 * 16.2)
note: 2 *0.5 = 1
By comparing the last answer with the answer of Tracy , we find that:
<span>Tracy’s answer will be correct because she made use of the fact that 2 (1/2) = 1 in her expression.
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</span>
(3)
the total surface area of this rectangular pyramid =
= area of two face with the height 52 + area of two face with the height 60 + area of the base
area of two face with the height 52 = 2 * 0.5 * 52 * 78 = 4,056 m²
area of two face with the height 60 = 2 * 0.5 * 60 * 50 = 3,000 m²
are of the base = 78 * 50 = 3900 m²
Total area = 4,056 + 3,000 + 3,900 = 10,956 m²
Assuming x is the hours each car has driven,
54x+56=52x+70
2x+56=70
2x=14
x=7
7 hours
52(7)+70
364+70
434 units (miles, yards, etc.)
The parenthesis's are only there to keep the two properties organized. The answer would be 6 since its just 9 minus 3 .
Answer:
![\frac{d(cos^{-1}x )}{dx} = \frac{-1}{\sqrt{1-x^2} }](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28cos%5E%7B-1%7Dx%20%29%7D%7Bdx%7D%20%3D%20%5Cfrac%7B-1%7D%7B%5Csqrt%7B1-x%5E2%7D%20%7D)
Step-by-step explanation:
Given the differential (d/dx)(cos−1(x)), to find the equivalent formula we will differentiate the inverse function using chain rule as shown below;
let;
![y = cos^{-1} x \\\\taking \ cos\ of\ both\ sides\\\\cosy = cos(cos^{-1} x)\\\\cosy = x\\\\x = cosy\\\\\frac{dx}{dy} = -siny\\](https://tex.z-dn.net/?f=y%20%3D%20cos%5E%7B-1%7D%20x%20%5C%5C%5C%5Ctaking%20%5C%20cos%5C%20of%5C%20both%5C%20sides%5C%5C%5C%5Ccosy%20%3D%20cos%28cos%5E%7B-1%7D%20x%29%5C%5C%5C%5Ccosy%20%3D%20x%5C%5C%5C%5Cx%20%3D%20cosy%5C%5C%5C%5C%5Cfrac%7Bdx%7D%7Bdy%7D%20%3D%20-siny%5C%5C)
![\frac{dy}{dx} = \frac{-1}{sin y} \\\\from\ trigonometry\ identity,\ sin^{2} x+cos^{2}x = 1\\sinx = \sqrt{1-cos^{2} x}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B-1%7D%7Bsin%20y%7D%20%20%5C%5C%5C%5Cfrom%5C%20trigonometry%5C%20identity%2C%5C%20sin%5E%7B2%7D%20x%2Bcos%5E%7B2%7Dx%20%3D%201%5C%5Csinx%20%3D%20%5Csqrt%7B1-cos%5E%7B2%7D%20x%7D)
Therefore;
![\frac{dy}{dx} = \frac{-1}{\sqrt{1-cos^{2}y } }](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B-1%7D%7B%5Csqrt%7B1-cos%5E%7B2%7Dy%20%7D%20%7D)
Since x = cos y from the above substitute;
![\frac{dy}{dx} = \frac{-1}{\sqrt{1-x^{2}} }](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7B-1%7D%7B%5Csqrt%7B1-x%5E%7B2%7D%7D%20%7D)
Hence,
gives the required proof