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3 years ago

5

What is the equation of the line that passes through the point (-8,5) and has a slope of o?

1 answer:

artcher [175]3 years ago

6 0

Answer: y = 5

Step-by-step explanation:

If the slope is 0, the graph of the line is a horizontal line (parallel to the x-axis). 0 slope means it is flat.

Here it passes through the y-axis at +5 and passes through <u>every</u> value of x.

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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

Write the next three terms of the following sequence 4,5,7,10,14

Law Incorporation [45]

4,5,7,10,14=19,25, 32

8 0

3 years ago

A car has mass 1500 kg and is traveling at a speed of 35 miles/hour. what is its kinetic energy in joules? (Be sure to convert m

photoshop1234 [79]

Answer:

Factor by which kinetic energy increase = 4 times

Step-by-step explanation:

Given,

Mass of the car, v1 = 1500 kg

initial speed of car = 35 miles/h

$=\dfrac{35\times 1609.34}{3600}\ m/s$

= 15.64 m/s

Initial kinetic energy of the car is given by,

$k_1\ =\ \dfrac{1}{2}.m.v_1^2$

$=\ \dfrac{1}{2}\times 1500\times (15.64)^2\ joule$

= 183606.46 J

Final velocity of car v2 = 70 miles/hour

$=\dfrac{70\times 1609.34}{3600}$

= 31.29 m/s

So, final kinetic energy of car is given by

$k_2\ =\ \dfrac{1}{2}.m.v_2^2$

$=\ \dfrac{1}{2}\times 1500\times (31.29)^2$

= 734425.84 J

Now, the ratio of final to initial kinetic energy can be given by,

$\dfrac{k_2}{k_1}=\ \dfrac{734425.84}{183606.46}$

$=>\ k_2\ =\ 4k_1$

Hence, the kinetic energy will increase by 4 times.

7 0

3 years ago

Together, Aanya and Krish can wash and wax Aanya's car in 50 minutes. Alone it takes Krish ten minutes more than Aanya. To the n

Leto [7]

Answer:

95 ; 105

Step-by-step explanation:

Let :

Aanya's time = x

Krish's time = x + 10

Together = 50

Hence

Aanya's rate = 1/x

Krish's rate = 1/(x + 10)

Together rate = 1/50

Hence,

1/x + 1/(x + 10) = 1 / 50

((x + 10) + x) / x(x + 10) = 1/ 50

(2x + 10)/ x² + 10x = 1/50

50(2x + 10) = x² + 10x

100x + 500 = x² + 10x

x² + 10x - 100x - 500 = 0

x² - 90x - 500 = 0

Using the quadratic equation solver :

x = 95.24 ; x = - 5.24

Time can't be negative

Hence, x = 95 (nearest minute)

It takes Aanya 95 minutes Alone a ND

Krishna (95 + 10) = 105 minutes alone

6 0

3 years ago

What is the operation used for the following terms: times, product, twice, etc.

Dafna1 [17]

Multiplication, hope I could help!

8 0

3 years ago

Read 2 more answers