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andriy [413]
3 years ago
5

Find the Pythagorean Triple (a,b,c) using the terms (x²-1, 2x, x²+1) for x > 0 when x =7.

Mathematics
1 answer:
yanalaym [24]3 years ago
4 0
1) 7^2-1=48
2x7=14
7^2+1=50
=(48,14,50)
2)10^2-6^2=64
2x10x6=120
10^2+6^2=136
=(64,120,136)
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Divide 1 pound by 3 which is approximately 7 divided by 3 the answer must be rounded up to a whole number
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3x-12/x^2+10x+24 ÷ x^2-16/x^2+6x.​
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Set up an equation that can be used to solve the problem. Solve the equation and determine the desired value.
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Answer:  $454.55

Step-by-step explanation:

Tito's weekly salary is always $ 500 per week.

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If we call x the dollar amount of all sales made by Tito in a week, then his income z(x) is given by the following equation:

z (x) = 0.55x + 500

We wish that z (x) = 750.

Then we equate the equation to 750 and solve for the variable x.

0.55x + 500 = 750\\\\0.55x + 500-500 = 750-500\\\\0.55x = 250

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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
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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}

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pishuonlain [190]

Answer:

2(5+9)_6

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