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skelet666 [1.2K]
2 years ago
5

How do you find the domain of f(x)=lnlnlnx

Mathematics
1 answer:
11Alexandr11 [23.1K]2 years ago
6 0
We have to find the domain of the function:
f ( x ) = ln ln ln x
x > 0
ln x > 0
x > 1
and finally: ln ( ln x ) > 0
ln x > 1
x > e
Answer: 
Domain is: x ∈ ( e , + ∞ ).
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Mell is twice as old as Jerry. If 16 is subtracted from Mell’s age and 16 is added to Jerry’s age, their ages will be equal. Wha
irga5000 [103]

Answer:


Step-by-step explanation:

Let J = Jerry's age

Let M = Mary's age = 2J

 

M - 16 = J + 16

 

Substitute 2J for Mary's age

 

2J - 16 = J + 16

2J - J - 16 = 16

J = 16 + 16

J = 32

 

Jerry is 32.

Mary is twice as old as Jerry.

Mary is 2(32) = 64

3 0
3 years ago
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3х – 6 =-2y<br> Find the slope of a line perpendicular to each given line
kondaur [170]

Answer:

Slope of the perpendicular line: 2/3

Step-by-step explanation:

-2y = 3x - 6

y = - 3/2x - - 6/2

y = -3/2 + 3

3 0
3 years ago
I have no idea!! Please help!!!
Ber [7]
The area of a traingle is \frac{1}{2}bh
So we need to know the base and the height
To work out the sides of a right angle triangle we can use a²+b²=c
c is the hypotenuse and a and b are the other sides
Because this triangle is isoceles we know that a and b are the same

a²+b²=c²
a²+a²=c²
2a²=c²
2a²=(x√6)²
2a²=(x√6)(x√6)
2a²=x²+x√6+x√6+6
2a²=x²+x√6+x√6+6
a²=\frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{2}
a=\sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}

A=\frac{1}{2}bh
A=\frac{1}{2}a*a
A=\frac{1}{2}×\sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}×\sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}
A=\frac{1}{2}×\frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{2}
A=\frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{4}
8 0
2 years ago
A company orders 29 boxed lunches from a deli for $324.80. If each boxed lunch
UkoKoshka [18]

Answer:

Each boxed lunch costs $11.20

Step-by-step explanation:

29 boxed lunches = $324.80

divide by 29 on each one

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