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2 years ago

I would like some help

Mathematics

1 answer:

Anettt [7]2 years ago

5 0

Answer:

Step-by-step explanation:

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I rlly need help pls help me

kifflom [539]

Answer:

This one ↓↓↓↓↓↓

Step-by-step explanation:

The picture attached is the reflection over the line y = 0

-Chetan K

4 0

2 years ago

Which is the best first step when solving the following system of equations? x + y = 3. 4 x minus y = 7. Multiply the first equa

Kryger [21]

Answer:

Multiply the first equation by 4.

Step-by-step explanation:

4(x+y=3) which gives

4x+4y=12

and

4x-y=7

Then subtract the two equations and the x variable will be eliminated.

6 0

3 years ago

Read 2 more answers

Which expression represents the following statement: x is no less than 0. x>0 x≤0 x≥0 x<0

ki77a [65]

Answer: x ≥ 0

Step-by-step explanation:

First, let's define the symbols used:

a < x (this means that a is strictly smaller than x)

a > x (this means that a is strictly larger than x)

a ≤ x (this means that a is smaller than or equal to x)

a ≥ x (this means that a is larger than or equal to x)

Now we have the statement "x is no less than 0"

Then x can be equal to zero, or larger than zero, but never smaller than zero.

Looking at the symbols above, we know that we need to use:

x ≥ 0

(this is equivalent to the statement)

7 0

3 years ago

Read 2 more answers

How do you solve this inequality; 1/3x>-2?

Kisachek [45]

1/3x>-2=1>3x.-2
=1>-6x
=-1/6>x

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)

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

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

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