Five times a complement of an angle exceeds twice its supplement by 9. Find the angle

Mathematics

1 answer:

kkurt [141]3 years ago

4 0

Answer:

Step-by-step explanation:

the answer is in the above image

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What is the slope of the line that passes through the points (1, 3) and (5, –2)?

Sergeu [11.5K]

m=-(5/4)

From left to right, (1,3) is first and then comes (5,-2). Always remember when finding slopes without equations, the rule is RISE over RUN, to the numerator and denominator, respectively.

The y value of the second coordinates becomes negative which is unlike the y value in the first coordinates, which means our slope is downward, meaning it has a negative sign in front.

In every slope, there’s a numerator, being the rise, and a denominator, being the run.

To find the rise, we must look at the y values. Starting at 3 going to -2 has a space of 5 units, making that our numerator.

To find the run, the first x value is 1 and the second is 5, making a space of 4, which is out denominator.

With these two numbers and the negative sign, we get -(5/4) as our slope.

7 0

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)

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