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Alchen [17]
3 years ago
15

Write an equation in point-slope form for the line through the given point with the given slope.

Mathematics
1 answer:
Lilit [14]3 years ago
7 0

The correct answer is D) y + 1 = 4/3(x - 9)

In order to find this, simply take the point and the slope and plug into the base form of point-slope form.

y - y1 = m(x - x1)

Use the slope for m and the point for (x1, y1)

y + 1 = 4/3(x - 9)

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#2. If someone could help me please
alina1380 [7]
The answer is D if you combine like terms
5 0
3 years ago
10 gallons and 11 quarts divided by 3
ivanzaharov [21]
4 quarts per gallon 
<span>10 gallons: 4x10 = 40 quarts. </span>
<span>Add the 11 quarts to get 51 quarts. </span>
<span>51 divided by 3 = 17 quarts, or 4 gallons + 1 quart.</span>
5 0
3 years ago
Read 2 more answers
What is the point-slope form of the equation of a line that passes through the point (1, 4) and has a slope of −3?
zmey [24]
So y=mx+b
m=slope

y=-3x+b
subsitutue 1 for x and 4 for  y
4=-3(1)+b
4=-3+b
add 3 to both sides
7=b

the equaiton is y=-3x+7
3 0
3 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

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
Arrange 3/5,5/8,5/6 and 7/4 in ascending order
melomori [17]

It’s already in ascending order.

3/5= .6

5/8= .625

5/6= .833

7/4= 1.75

6 0
2 years ago
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