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2 years ago

Can someone please help me on this question I will give brainliest

Mathematics

1 answer:

xz_007 [3.2K]2 years ago

4 0

Answer:

A) 3 in

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Geometry

Surface Area of a Sphere: SA = 4πr²
Diameter: d = 2r

Step-by-step explanation:

Step 1: Define

SA = 23 in²

Step 2: Find r

Substitute [SAS]: 23 in² = 4πr²
Isolate r term: 23 in²/(4π) = r²
Isolate r: √[23 in²/(4π)] = r
Rewrite: r = √[23 in²/(4π)]
Evaluate: r = 1.35288 in

Step 3: Find d

Substitute [D]: d = 2(1.35288 in)
Multiply: d = 2.70576 in
Round: d ≈ 3 in

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Answer:

$y(t) = y_o e^{\beta t}$

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

$y(t) = e^{\beta t + c}$

=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

$y(t) = y_o e^{\beta t}$

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

$\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

5 0

2 years ago

If g(x) = x2 + 2, find g(3). 9 8 11 6

cestrela7 [59]

Answer

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To proof

The function given in the question is

g(x) = x² + 2

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put x = 3 in the g(x) = x² + 2

than it becomes

g(3) = 3² + 2

solving the above

we get

g(3) = 9 + 2

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Thus g(3) = 11 and option (c) is correct .

Hence proved

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V125BC [204]

Answer:

Step-by-step explanation:

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Vitek1552 [10]

Hey there!!

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