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Tatiana [17]
3 years ago
7

What are the solutions to lx + 2) = -13?

Mathematics
1 answer:
Masja [62]3 years ago
8 0

Answer:

B

Step-by-step explanation:

Because -15 + 2 = -13

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Yolanda buys
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Answer:

The unit price is 4.15 dollars per gallon

Step-by-step explanation:

- Yolanda buys 5.8 gallons of gas for $24.07

- We need to find the unit price in dollars per gallon

- Unit price means the price of each gallon

∵ The number of gallons = 5.8 gallons

∵ The price of the gallons = $24.07

∵ The unit price = the amount of money ÷ number of gallons

∴ The unit price = 24.07 ÷ 5.8

∴ The unit price = 4.15 dollars per gallon

* <em>The unit price is 4.15 dollars per gallon</em>

3 0
3 years ago
a classroom has a length of 20 feet and a width of 30 feet. The headmaster decided that tiles will look good in that class. if e
iogann1982 [59]
The area of the room must equal the sum of the areas of the tiles...

20*30=n(2*3)

600=6n

n=100

So 100 tiles are needed.
8 0
3 years ago
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PLEASE ANSWER THE QUESTION ON THE SCREENSHOT ASAP!
mars1129 [50]
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4 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

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
4=(1/2)^× solve for x
cricket20 [7]
4=\left(\frac{1}{2}\right)^x\\&#10;2^2=2^{-x}\\&#10;x=-2&#10;
4 0
3 years ago
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