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RoseWind [281]
3 years ago
13

1+4=5 2+5=12 3+6=21 5+8=?

Mathematics
1 answer:
Korvikt [17]3 years ago
7 0
As I see it, every addition also adds the results of the last so, 34
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19 - (- 41)<br> plzzzzzz help fast!
leonid [27]

Answer:

60

Step-by-step explanation:

Is because when you have two negative -- signs it became positive

19 + 41 = 60

3 0
3 years ago
Read 2 more answers
Explain why is 1/5 bigger than 0.15
Nitella [24]
1/5 is greater because it equals 0.2

0.2 has an imaginary 0 at the end, to make it 0.20

.20 > 0.15

Therefore, 1/5 is greater than 0.15
8 0
3 years ago
Solve 7x+6&lt;3(x - 2).
lawyer [7]

Answer:

x<−3

Step-by-step explanation:

hope this works

4 0
3 years ago
Please please help me
Anarel [89]

so you are going to use the Pythagorean theorem on the smaller triangle to find the connecting side's length

{5}^{2}  = {3}^{2} ( {b}^{2} ) \\ b = 4

now you can find the bigger triangle's hypotenuse

{6}^{2}  +  {4}^{2}  =  {c}^{2}  \\

c=4

8 0
3 years ago
Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be
Travka [436]

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

dA/dt = in flow - out flow

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

dA/dt = L - Ak

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

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

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}

4 0
3 years ago
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