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3 years ago

Which function represents a vertical stretch of an exponential function

1 answer:

7 0

Vertical stretch means the graph stretches upward, rising more rapidly than the original, the graph is steeper than the original.

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Find all the square roots of x2 ≡ 53 (mod 77)

Nikolay [14]

Answer:

$x=\pm\sqrt{77n+53}$

Step-by-step explanation:

We have been given an equivalence equation $x^2\equiv 53\text{ (mod } 77)$ . We are asked to find all the square root of the given equivalence equation.

Upon converting our given equivalence equation into an equation, we will get:

$x^2-53=77n$

Add 53 on both sides:

$x^2-53+53=77n+53$

$x^2=77n+53$

Take square root of both sides:

$x=\pm\sqrt{77n+53}$

Therefore, the square root for our given equation would be $x=\pm\sqrt{77n+53}$ .

3 0

3 years ago

Designing a Sprinkler System

Bogdan [553]

Answer: Its B

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Please find the missing number and please don’t refer me to a link thank youu

svlad2 [7]

Answer:

i don't know the answer BUT you can go to khan academy or tutor.com. Both can help :) good luck btw

Step-by-step explanation:

7 0

2 years ago

Factorise fully 77x-33x^2

lesya [120]

Answer:

11x(7 - 3x)

Step-by-step explanation:

77x - 33x² ← factor out 11x from each term

= 11x(7 - 3x)

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)

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

$A = \frac{L}{k} - \frac{L}{k} e^{kt}$

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