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Can anyone help me with question number 3

ira [324]

The answer is: There are 595 calories in a banana split

To find this we first need to set up our equation:

$80= \frac{1}{7}x-5$ , where x= calories in a banana split

$85= \frac{1}{7}x$ ----------------Add 5 to both sides

595=x-------------------Multiply 7 to both sides to get rid of the fraction

So, there are 595 calories in a banana split

Hope this helps! :)

7 0

3 years ago

In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+3y=8te^−t+6e^−t−(9t+6)

Luden [163]

We're given the ODE,

y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) - (9t + 6)

(where I denote exp(x) = eˣ )

First determine the characteristic solution:

y'' + 4y' + 3y = 0

has characteristic equation

r ² + 4r + 3 = (r + 1) (r + 3) = 0

with roots at r = -1 and r = -3, so the characteristic solution is

y = C₁ exp(-t ) + C₂ exp(-3t )

For the non-homogeneous equation, assume two ansatz solutions

y₁ = (at ² + bt + c) exp(-t )

and

y₂ = at + b

• y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) … … … [1]

Compute the derivatives of y₁ :

y₁ = (at ² + bt + c) exp(-t )

y₁' = (2at + b) exp(-t ) - (at ² + bt + c) exp(-t )

… = (-at ² + (2a - b) t + b - c) exp(-t )

y₁'' = (-2at + 2a - b) exp(-t ) - (-at ² + (2a - b) t + b - c) exp(-t )

… = (at ² + (b - 4a) t + 2a - 2b + c) exp(-t )

Substitute them into the ODE [1] to get

→ [(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c)] exp(-t ) = 8t exp(-t ) + 6 exp(-t )

(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c) = 8t + 6

4at + 2a + 2b = 8t + 6

→ 4a = 8 and 2a + 2b = 6

→ a = 2 and b = 1

→ y₁ = (2t ² + t ) exp(-t )

(Note that we don't find out anything about c, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-t ) anyway.)

• y'' + 4y' + 3y = -(9t + 6) … … … [2]

Compute the derivatives of y₂ :

y₂ = at + b

y₂' = a

y₂'' = 0

Substitute these into [2] :

4a + 3 (at + b) = -9t - 6

3at + 4a + 3b = -9t - 6

→ 3a = -9 and 4a + 3b = -6

→ a = -3 and b = 2

→ y₂ = -3t + 2

Then the general solution to the original ODE is

y(t) = C₁ exp(-t ) + C₂ exp(-3t ) + (2t ² + t ) exp(-t ) - 3t + 2

Use the initial conditions y (0) = 2 and y' (0) = 2 to solve for C₁ and C₂ :

y (0) = C₁ + C₂ + 2 = 2

→ C₁ + C₂ = 0 … … … [3]

y'(t) = -C₁ exp(-t ) - 3C₂ exp(-3t ) + (-2t ² + 3t + 1) exp(-t ) - 3

y' (0) = -C₁ - 3C₂ + 1 - 3 = 2

→ C₁ + 3C₂ = -4 … … … [4]

Solve equations [3] and [4] to get C₁ = 2 and C₂ = -2. Then the particular solution to the initial value problem is

y(t) = -2 exp(-3t ) + (2t ² + t + 2) exp(-t ) - 3t + 2

7 0

2 years ago

Type the integer that makes the following addition sentence true:

enot [183]

Answer:

93

Step-by-step explanation:

Add 36 to both sides of the equation:

36 -36 + ___ = 36 +57

0 + ___ = 93

___ = 93

The integer 93 goes in the blank.

3 0

2 years ago

Me ayuden por favor es para Hoy ✌✌

DIA [1.3K]

ninguna de las preguntas se puede dividir, pero puede simplificar la última.

Lo siento pero, Espero que esto sea de ayuda!

8 0

2 years ago

In △DEF, DF = 17 and m∠F=32. Find DF to the nearest tenth.

SSSSS [86.1K]

FE=14.42
ED=9.01
angle D=58
hope this helps

8 0

3 years ago