Answer:
Step-by-step explanation:
You know how subtraction is the <em>opposite of addition </em>and division is the <em>opposite of multiplication</em>? A logarithm is the <em>opposite of an exponent</em>. You know how you can rewrite the equation 3 + 2 = 5 as 5 - 3 = 2, or the equation 3 × 2 = 6 as 6 ÷ 3 = 2? This is really useful when one of those numbers on the left is unknown. 3 + _ = 8 can be rewritten as 8 - 3 = _, 4 × _ = 12 can be rewritten as 12 ÷ 4 = _. We get all our knowns on one side and our unknown by itself on the other, and the rest is computation.
We know that 
; as a logarithm, the <em>exponent</em> gets moved to its own side of the equation, and we write the equation like this: 
, which you read as "the logarithm base 3 of 9 is 2." You could also read it as "the power you need to raise 3 to to get 9 is 2."
One historical quirk: because we use the decimal system, it's assumed that an expression like 
uses <em>base 10</em>, and you'd interpret it as "What power do I raise 10 to to get 1000?"
The expression 
means "the power you need to raise 10 to to get 100 is x," or, rearranging: "10 to the x is equal to 100," which in symbols is .
(If we wanted to, we could also solve this: 
, so 
)
Hello :
all n in N ; n(n+1)(n+2) = 3a a in N or : <span>≡ 0 (mod 3)
1 ) n </span><span>≡ 0 ( mod 3)...(1)
n+1 </span>≡ 1 ( mod 3)...(2)
n+2 ≡ 2 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 0×1×2 ( mod 3) : ≡ 0 (mod 3)
2) n ≡ 1 ( mod 3)...(1)
n+1 ≡ 2 ( mod 3)...(2)
n+2 ≡ 3 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 1×2 × 3 ( mod 3) : ≡ 0 (mod 3) , 6≡ 0 (mod)
3) n ≡ 2 ( mod 3)...(1)
n+1 ≡ 3 ( mod 3)...(2)
n+2 ≡ 4 ( mod 3)...(3)
by (1), (2), (3) : n(n+1)(n+2) ≡ 2×3 × 4 ( mod 3) : ≡ 0 (mod 3) , 24≡ 0 (mod3)
Ok so if 1200 total pens and 60 were taken that's a ratio of 1200/60=20
out of 60pens 4 didn't work, again ratio is 4/60=1/15 so about 1/15 of the pens didn't work.
so we have 1200pens * 1/15broken ratio = 80 pens or we can say
20*4=80pens won't work from 20 groups of 60 and 4 out of 60 not working is 20*4=80 not working from total batch of 1200.
hope this helps you some! thank you!!!!!!!!!
Answer:
Use trig ratios to find unknown sides in right triangles.
Step-by-step explanation:
Hello again lol. I would think it would be two rooms don’t count me on this one but I think it is two
