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Ksenya-84 [330]
3 years ago
8

Question: Find the equivalent number. (I will do brainlisit )

Mathematics
2 answers:
11111nata11111 [884]3 years ago
7 0

you're answer ist going to Be B.9

3^4

3×3×3×3

81

3^2

3×3

9

81/9 = 9

vodomira [7]3 years ago
3 0

Answer:

81/9

9/1

9 is your answer ☺️☺️☺️

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What does a equal a=l
Akimi4 [234]
It says right there in your question a=l. So, a=l.

Now we have concluded that a=l. a equals l.
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Calculate the perimeter and area of this shape.<br> Perimeter =
konstantin123 [22]
<h3><u>Answer</u><u>:</u></h3>

Perimeter → 25cm

Area → 25.5cm²

<h3><u>Explanation</u>:</h3>

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7 0
1 year ago
Determine the equivalent system for the given system of equations: <br> 5x − 3y = 6 <br> x + y = 2
Arada [10]

Answer:

<h2>x = 1.5 and y = 0.5 → (1.5, 0.5)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}5x-3y=6\\x+y=2&\text{multiply both sides by 3}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}5x-3y=6\\3x+3y=6\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad8x=12\qquad\text{divide both sides by 8}\\.\qquad x=1.5\\\\\text{put the value of x to the second equation:}\\\\1.5+y=2\qquad\text{subtract 1.5  from both sides}\\y=0.5

3 0
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The element Unobtainium has a half-life of 3 years. Let M(t) be the mass of Unobtainium at time t starting with an initial amoun
Sphinxa [80]

Answer:

Part 1

a) M(t) = 14 e⁻⁰•²³¹ᵗ

b) 11.4 years

Part 2

a) The differential equation for T is

(dT/dt) = -k(T - T∞)

(dT/dt) = -k(T - 70)

And the solution of the differential equation

T(t) = 70 + 100e⁻⁰•¹²⁰⁴ᵗ

b) The additional minutes it will take for the temperature of the coffee to reach 80°C after reaching 100°C = 9.1 minutes.

Step-by-step explanation:

Part 1

Radioactive reactions always follow a first order reaction dynamic

Let the initial concentration of Unobtainium be M₀ and the concentration at any time be M

(dM/dt) = -kM (Minus sign because it's a rate of reduction)

(dM/dt) = -kM

(dM/M) = -kdt

 ∫ (dM/M) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from M₀ to M and the Right hand side from 0 to t.

We obtain

In (M/M₀) = -kt

(M/M₀) = e⁻ᵏᵗ

M(t) = M₀ e⁻ᵏᵗ

Although, we can obtain k from the information on half life.

For a first order reaction, the rate constant (k) and the half life (T(1/2)) are related thus

T(1/2) = (In2)/k

T(1/2) = 3 years

k = (In 2)/3 = 0.231 /year.

And M₀ = 14 kg from the question, M(t) becomes

M(t) = 14 e⁻⁰•²³¹ᵗ

b) What is t, when M(t) = 1 kg

M(t) = 14 e⁻⁰•²³¹ᵗ

1 = 14 e⁻⁰•²³¹ᵗ

e⁻⁰•²³¹ᵗ = (1/14) = 0.07143

-0.231t = In (0.07143) = - 2.639

t = 2.639/0.231 = 11.4 years

Part 2

a) Let T be the temperature of the coffee at any time

T∞ be the temperature of the room = 73°C

T₀ be the initial temperature of the coffee = 170°C

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the coffee = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

T(t) = T∞ + (T₀ - T∞)e⁻ᵏᵗ

Inputting the known variables

T(t) = 70 + (170 - 70) e⁻ᵏᵗ

T(t) = 70 + 100e⁻ᵏᵗ

But it is given that, at t = 10, T = 100°C

100 = 70 + 100e⁻¹⁰ᵏ

100e⁻¹⁰ᵏ = 30

e⁻¹⁰ᵏ = 0.3

-10k = In (0.3) = 1.204

K = 0.1204

T(t) = 70 + 100e⁻⁰•¹²⁰⁴ᵗ

b) T(t) = 70 + 100e⁻⁰•¹²⁰⁴ᵗ

We first calculate the time it takes to get 80°C

80 = 70 + 100e⁻⁰•¹²⁰⁴ᵗ

100e⁻⁰•¹²⁰⁴ᵗ = 10

e⁻⁰•¹²⁰⁴ᵗ = 0.1

-0.1204t = In (0.1) = -2.303

t = (2.303/0.1204) = 19.1 minutes.

From the time it reached 100°C, i.e. t = 10 minutes, 19.1 minutes is 9.1 minutes extra.

7 0
3 years ago
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