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OverLord2011 [107]
3 years ago
14

watermelon seed is 4mm long. How far would a line of 120 watermelon seeds stretch? Write your answer in meters.

Mathematics
1 answer:
rodikova [14]3 years ago
3 0
120 seeds would be 120 \cdot 4 = 480 milimeters. This is also \boxed{0.48} meters (since 1000mm is 1 m).
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Multiply 2(x+3)/x(x-1)*4x(x+2)/10(x+3)
KIM [24]

Answer:

The expression simplifies to \frac{4(x+2)}{5(x-1)}.

Step-by-step explanation:

The expression

\frac{2(x+3)}{x(x-1)} *\frac{4x(x+2)}{10(x+3)}

can be rearranged and written as

\frac{8x(x+3)(x+2)}{10x(x-1)(x+3)}.

In this form the (x+3) terms in the numerator and in the denominator cancel to give

\frac{8x(x+2)}{10x(x-1)}.

The x's are present both in the numerator and in the denominator, so they also cancel, and the fraction \frac{8}{10} simplifies to \frac{4}{5}, so finally our expression becomes:

\boxed{\frac{4(x+2)}{5(x-1)}}

Which is our answer:)

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Gwar [14]
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3 years ago
5 pounds is how many ounces
chubhunter [2.5K]

Answer:

80 ounces

Step-by-step explanation:

There are 16 ounces in a pound. So to convert from pounds to ounces, you multiply 5 by 16, which is 80.

4 0
3 years ago
What is 12925 rounded to the nearest ten thousand
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Read 2 more answers
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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