Tan ( a + b ) = [ tan a + tan b ] / [ 1 - (tan a)*(tan b) ];
let be a = 2x and b = x;
=> tan 3x = [ tan 2x + tan x ] / [ 1 - (tan 2x)*(tan x) ] => (tan 3x)*[ 1 - (tan 2x)*(tan x) ] =
tan 2x + tan x => tan3x - tan 3xtan 2xtanx = tan 2x + tan x => <span> tan 3x−tan 2x−tanx = tan 3xtan 2xtanx.</span><span />
Answer:
I think you solved it my g
Answer:

Step-by-step explanation:
The change of form formula (exponential to logarithm) is:

Looking at the problem given, we can simply write the logarithmic form as:

This is the answer.
Answer:
What is the probability that a randomly selected family owns a cat? 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat? 82.4%
Step-by-step explanation: We can use a Venn (attached) diagram to describe this situation:
Imagine a community of 100 families (we can assum a number, because in the end, it does not matter)
So, 30% of the families own a dog = .30*100 = 30
20% of the families that own a dog also own a cat = 0.2*30 = 6
34% of all the families own a cat = 0.34*100 = 34
Dogs and cats: 6
Only dogs: 30 - 6 = 24
Only cats: 34 - 6 = 28
Not cat and dogs: 24+6+28 = 58; 100 - 58 = 42
What is the probability that a randomly selected family owns a cat?
34/100 = 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?
A = doesn't own a dog
B = owns a cat
P(A|B) = P(A∩B)/P(B) = 28/34 = 82.4%
2x+10=0
subtract 10 from both sides
2x=(-10)
divide 2 from both sides
x=-5