What is the product of 36 and 80

Mathematics

1 answer:

Nitella [24]3 years ago

5 0

The product of 36 and 80 is 2880.

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How to estimate 184÷9 in multiplication

FrozenT [24]

Do you mean checking your work? In that case it would be 9*20.

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3 years ago

Needing help to find the inequality

tiny-mole [99]

http://goblues.org/faculty/grantm/files/2015/12/Practice-Test-3.7-Solutions.pdf

I think it's question number three. Sorry if the link dosen't work.

4 0

2 years ago

Sara left a bin outside in her garden to collect rainwater. She notices that 1/8 gallon of water fills 2/3 of the bin. Write and

Tpy6a [65]

Answer: 3/16 gallons of water

Step-by-step explanation:

With any fraction, if you multiply it by its reciprocal, it equals 1. (ex: 3/4*4/3=1, 1/2*2/1=1, 71/256*256/71=1) Thus, 2/3*3/2=1. Because 2/3 of the bin=1/8 gallon, simply multiply both sides of the equation by 3/2 and you will get that 1 bin of water=0.1875 gallons of water, or 3/16 gallons of water.

Hope it helps <3

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2 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)

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