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2 years ago

6 of 25

Mathematics

2 answers:

grin007 [14]2 years ago

6 0

$9.513 per meal. hope this helps :)

olasank [31]2 years ago

3 0

9.5139 because you could divide 599.38 by 63 to get 9.5139

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3x+6<25,x=8 true or false

Liono4ka [1.6K]

That is false

If x=8 then the whole thing is 24+6 which is 30

so 30<25 is false

6 0

3 years ago

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Find the selling price. Cost to store: $90 Markup: 60%

Arturiano [62]

$135
60% markup means 60 percent of 90 dollars added back into the price
45 is 60 percent of 90 dollars so you would add that to $90

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3 years ago

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Which two numbers on the number line have an absolute value of 1.75? Select the location of both numbers on the number line

hram777 [196]

Solution

To find Which two numbers on the number line have sn

absolute value of 1.75 .

Absolute value means " how far a number is from

zero " .

"1.75" is 1.75 away from zero .

and " -1.75" is also 1.75 away from zero .

i.e

| 0 - 1.75 | = 1.75

| 1.75 - 0 | = 1.75

So , the absolute value of "1.75" is 1.75

similarly , absolute value of "1.75" is 1.75 .

Hence , the two numbers are -1.75 and 1.75 .

8 0

3 years ago

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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)

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