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Readme [11.4K]
3 years ago
10

Simplify the expression: show all ur steps/ wrk PLEASE,1+2x3-18/3+(5(2+2)]

Mathematics
2 answers:
Maurinko [17]3 years ago
4 0
Ok let's do this*pops knuckles*



let's add the first side

1 + 2 \times 3 - 18
order of operations demands that we do multiplication first(P.E.M.D.A.S)

so
2 \times 3 = 6
now we can add the 1 from
1 + (6) - 18
into our six


6 + 1 = 7
now we substract

(7) - 18 =  - 11
now we have
- 11  \div 3 + (5(2 + 2))
now we focus on the
other side
3 + (5(2 + 2))
since 5 is in paranthesis we use something called the distributive property. it mean we will multiply the 5 by both of the 2.
(5(2 + 2)) = (((5 \times 2) + (5 \times 2) = 20
now we have
3  + 20 = 23

(the three is from the original equation)

now we plug in our final numbers

- 11 \div 23
or
\frac{ - 11}{23}
I hope this helps
Oxana [17]3 years ago
4 0
1+2x3-18/3+[5(2+2)]
1+2*3-18/3+[5(4)]
1+2*3-18/3+[5*4]
1+2*3-18/3+[20]
1+6-18/3+20
1+6-6+20
7-6+20
1+20
=21

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Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

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Since litter falls at a constant rate of L  grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

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Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C'      (C' = kC)

taking exponents of both sides, we have

L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt}      (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k}  - \frac{C"}{k} e^{kt}

When t = 0, A(0) = 0 (since the forest floor is initially clear)

A = \frac{L}{k}  - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k}  - \frac{C"}{k} e^{0}\\\frac{L}{k}  = \frac{C"}{k} \\C" = L

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So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

when t = 0(at initial time), the initial value of D =

D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}

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