II has a radius of 1.47 inches and weighs 11 grams per cubic inch Which of the following

Find an equation for the plane through the points p=(0,0,0), q=(0,1,0), and r=(1,2,3)

jek_recluse [69]

Hello,

Assume Ax+By+Cz+D=0 the plane 's equation.

p=(0,0,0)==>A*0+B*0+C*0+D=0==>D=0

q=(0,1,0)==>A*0+B*1+C*0+0=0==>B=0

r=(1,2,3)==>A*1+0*2+C*3+0=0==>A=-3C
Let C=1==>A=-3

An equation of the plane is -3x+z=0

6 0

3 years ago

What is the value of k used to find the coordinates of a point that partitions a segment into a ratio of 5:3. Please Explain.

Dafna11 [192]

Answer:

The value of k is 5/8

Step-by-step explanation:

The value of k is found by dividing the numerator of the original ratio, 5, by the sum of the numerator and denominator of the ratio

When finding a point, P, to partition a line segment AB into the ratio a/b, we find a ratio c = a / (a + b)

According to this formula we find the value of k.

k = a/(a+b)

where a = 5

b = 3

Now plug the values in the formula:

k = 5/(5+3)

k = 5/8 ....

4 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)

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