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masya89 [10]
3 years ago
11

One of the factors of 6x3 − 864x is A. 4 B. X^2 C. X + 12 D. X - 8

Mathematics
2 answers:
erastova [34]3 years ago
8 0
The answer is C. X+12
Anna71 [15]3 years ago
4 0

Consider the expression 6x^3-864x. This expression consists of two terms: 6x^3 and -864x.

These two terms have common factors 6 and x, because:

  • 6x^3=6x\cdot x^2;
  • -864x=6x\cdot (-144).

Then

6x^3-864x=6x(x^2-144).

You know that 144=12^2. Use the formula for the difference of squares:

x^2-144=x^2-12^2=(x-12)(x+12).

Therefore,

6x^3-864=6x(x-12)(x+12).

Answer: correct choice is C.

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The populations of termites and spiders in a certain house are growing exponentially. The house contains 120 termites the day yo
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Answer:

in order to triple the inicial population of spiders, will take 50395 days

Step-by-step explanation:

we can define the termite population function as T(t) and the one for spiders as S(t) , where t represents time measured in days

since both have and exponencial growth

T(t)= a*e^(b*t)

S(t)= c*e^(d*t)

1) when the day the person moves in , t=0 and T(0)= 120 termites

T(0) = a*e^(b*0) = a = 120

2) after 4 days , t=4 and  the house contains T(4) = 210 termites

T(4)= 120*e^(b*4) = 210 → 4*b = ln (210/120) → b = (1/4)* ln(210/120)= 0.14

therefore

T(t) = 120*e^(0.14*t)

3) 3 days after moving in , t=3, there were T(3) = 120*e^(0.14*3)=182.63≈ 182 termites . The number of spiders is half of the number of termites → S(3) = T(3) * 1 spider/ 2 termites  =91.31 spiders ≈ 91 spiders

4) after 8 days of moving in , t=8, there were T(8) = 120*e^(0.14*8)=367.78≈ 368 termites . The number of spiders is 0.25 times the number of termites → S(8) = T(8) * 1 spider/ 4 termites =91.94 spiders   ≈ 92 spiders

from

S(t)= c*e^(d*t) → d = ln [S (tb)/S (ta) ] / (tb-ta)

therefore d = ln [ S(8)/S(3) ] / (8 - 3 ) = 2.18*10^-5

in order to triple the initial population

S(t3) = 3 *S(0) = 3*[c*e^(d*0)] = 3*c

S(t3) = c*e^(d*t3) = 3* c → t3 = ln(3) / d = 50395 days

5 0
3 years ago
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