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gladu [14]
3 years ago
11

Which situation represents a multiplicative relationship? A) Kevin is 8 years older than his brother Sean. B) For every cat at t

he pet store, there are 2 dogs. C) A buy one get one free sale on sweaters at the clothing store. D) James always keeps $100 more in his savings than his checking account.
Mathematics
1 answer:
Lunna [17]3 years ago
6 0

Answer:

Option B represents a multiplicative relationship.

Step-by-step explanation:

We are required to find the situation that represents a 'multiplicative relationship'.

According to the options, we have,

A) Kevin is 8 years older than his brother Sean.

This gives us the relation that, Kevin's Age = 8 + Sean's Age.

Thus, this is not a multiplicative relationship.

B) For every cat at the pet store, there are 2 dogs.

We get that, for 1 cat, there are two dogs.

So, Number of dogs = 2 × Number of cats.

Thus, this is a multiplicative relationship.

C) A buy one get one free sale on sweaters at the clothing store.

So, during the sale a person can buy 1 + 1 i.e. 2 sweaters.

Thus, this is not a multiplicative relationship.

D) James always keeps $100 more in his savings than his checking account.

This gives us the relation that, Savings balance = 100 + Checking balance

So, this is not a multiplicative relationship.

Hence, option B represents a multiplicative relationship.

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Answer:

y(x)=sin(2x)+2cos(2x)

Step-by-step explanation:

y''+4y=0

This is a homogeneous linear equation. So, assume a solution will be proportional to:

e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda

Now, substitute y(x)=e^{\lambda x} into the differential equation:

\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0

Using the characteristic equation:

\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0

Factor out e^{\lambda x}

e^{\lambda x}(\lambda ^2 +4) =0

Where:

e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda

Therefore the zeros must come from the polynomial:

\lambda^2+4 =0

Solving for \lambda:

\lambda =\pm2i

These roots give the next solutions:

y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}

Where c_1 and c_2 are arbitrary constants. Now, the general solution is the sum of the previous solutions:

y(x)=c_1 e^{2ix} +c_2 e^{-2ix}

Using Euler's identity:

e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)

y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\

Redefine:

i(c_1-c_2)=c_1\\\\c_1+c_2=c_2

Since these are arbitrary constants

y(x)=c_1sin(2x)+c_2cos(2x)

Now, let's find its derivative in order to find c_1 and c_2

y'(x)=2c_1 cos(2x)-2c_2sin(2x)

Evaluating    y(0)=2 :

y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2

Evaluating     y'(0)=2 :

y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1

Finally, the solution is given by:

y(x)=sin(2x)+2cos(2x)

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