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OLEGan [10]
3 years ago
10

What is the value of 7-3+(6-2•4)

Mathematics
2 answers:
Valentin [98]3 years ago
8 0
7-3+(6-2•4)
7-3+(6-8)
7-3+(-2)
7-3-2
4-2
2
defon3 years ago
7 0

Answer:

2

Step-by-step explanation:

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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
zysi [14]

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

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Find the missing side
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Answer:

$2.25

Step-by-step explanation:

10*0.10= $1.00

5*0.25= $1.25

1+1.25= $2.25

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(05.03 LC) What is the initial value of the function represented by this graph? (1 point) ​
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Step-by-step explanation:

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You have 2/3 of sugar how muck sugar will you use if you makes 6 batches of cookies
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1 batch = 2/3

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