Answer:
Step-by-step explanation:
Given two points are (-1,7) and (7,7).
We are told to find a point which is in between them at a ratio of 6:5.
Formula for calculating a point at distance of ratio m:n is given by
where 
are the inital and final points.
also m:n=6:5 ,
by substituting all the values we get,
(f o g)(x)
g(x) = x + 2
Since it’s telling you to use g(x), you use (x + 2) as your new x value for function f.
f(x + 2) = 5(x + 2) + 5
= 5x + 10 + 5
= 5x + 15
The correct option is "d".
Given that L = 0.8T²
length of pendulum = 30ft
L= 0.8T²
30 = 0.8T²
T² = 30 / 0.8
T² = 37.5
T = √37.5 = 6.1 seconds
<span>So, 6.1 is the closest to the period in seconds for a pendulum that is 30 ft long.</span>
Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
![\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2%2B%28%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot6%29%5E2%3D36%2B27%3D63%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B63%7D%20%5Cend%7Bgathered%7D)
so, the lateral area =
![6\cdot\frac{1}{2}\cdot6\cdot\sqrt[]{63}=18\sqrt[]{63}](https://tex.z-dn.net/?f=6%5Ccdot%5Cfrac%7B1%7D%7B2%7D%5Ccdot6%5Ccdot%5Csqrt%5B%5D%7B63%7D%3D18%5Csqrt%5B%5D%7B63%7D)
A, using the formula length= Area: width
and we have the length= 3535:2525= 1,4 m or this number as a fraction 14/10=7/5
But the length is smaller than the width?
b, I think you type wrongly the problem
