It is known that the generatrix of a cone is 13 cm long and 12 cm high The answer is 400 PI out of nine

It is known that the generatrix of a cone is 13 cm long and 12 cm high The answer is 400 PI out of nine

Let H be the cone height, l be the generatrix length, a be the base radius, R be the inscribed sphere radius, R be the circumscribed sphere radius, then H & nbsp; = & nbsp; 12 & nbsp; cm, & nbsp; L & nbsp; = & nbsp; 13 & nbsp; CMA & nbsp; = & nbsp; sqrt (13 ^ 2-12 ^ 2) & nbsp; = & nbsp; 5 can be seen from the figure & nbsp; X ^ 2 + R ^ 2 & nbsp; = & nbsp; y ^ 2 and x value

The generatrix of a straight cone is 10 cm long and 8 cm high. The volume of the inscribed sphere of the cone is calculated

First, the radius of the inscribed sphere R is obtained
Bottom diameter of cone = 2 √ (10 & sup2; - 8 & sup2;) = 2 * 6 = 12cm
r=12*8/(10+10+12)=3cm
The volume of the inscribed sphere is 4R & sup3 / 3 = 4 * 27 / 3 = 36cm & sup3;

Given that the generatrix of a cone is 10 cm long and 8 cm high, calculate the surface area of the inscribed sphere of the cone

S = 4 π r squared
R = root 10-8 = 6
So s = 144 π
I'm not sure

The generatrix of a cone is 10 in length and 8 in height

As shown in the figure, the cone cross section, let the radius of the inscribed sphere be x, from the known figure and line relationship, from the similar triangle, (8-x) / 10 = x / 610x = 48-6x16x = 48x = 3, the radius of the inscribed sphere is found, the volume will be calculated

The inverse function of F (x) = log (2) x + 1

f（Ⅹ）=log(2)X+1
The range is r
Let y = log2 [x] + 1
Then log2 [x] = Y-1
x=2^(y-1)
The inverse function is
Y-2 ^ (x-1) is defined as R
So the inverse function is
F (x) = 2 ^ (x-1) is defined as R

Inverse function of 1-log with 1 / 2 as base x (x > = 1)

Step 1: find the definition field of the original function. The definition field of the original function has been given, which is x ≥ 1. Step 2: exchange X and y, solve y. exchange X and y of the original function y = 1-log (1 / 2, x) (x ≥ 1), get x = 1-log (1 / 2, y) (Y ≥ 1). Take it as an equation about y, solve y, get y = (1 / 2) ^ (- x + 1) (Y ≥ 1)

Let y = 1 + log a (x + 3)

y=1+loga(x+3)
loga(x+3)=y-1
a^(y-1)=x+3
x=a^(y-1)-3
The inverse function is:
y=a^(x-1)-3

The inverse function of the function y = 1 + Log & # 189; X is
A.y=2^x
B.y=(½)^x
C.y=2^1-x
D.y=2^x-1

y=1+log(1/2)(x)
y-1=log(1/2)(x)
x=(1/2)^(y-1)
y=(1/2)^(x-1)=2^(1-x)
C

Finding the inverse function of function y = log x + 1 / X-1

Y=lg（ x+1）/（x-1）
Inverse function x = LG (y + 1) / (Y-1)
10^x=（y+1）/（y-1）
10^x=（y-1+2）/（y-1）
10^x=1+2/（y-1）
2/（y-1）=1+10^x
y-1=2/（10^x-1）
y=1+2/（10^x-1）
There's no base. Should it be LG?

What is the inverse function of y = log a (x + (x ^ 2-1) ^ (1 / 2))

Y = [(a ^ (2 * x)) + 1] / 2 * a ^ x