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2 years ago

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Renee is buying Halloween candy. She wants to buy candy bars and lollipops and spend no more than $28. Each candy bar costs $0.4

5 and each lollipop costs $0.25. Which inequality represents the possible combinations of x, the number of candy bars, and y, the number of lollipops Renee can buy?

1 answer:

aivan3 [116]2 years ago

6 0

The inequality that represents the possible combinations of candy bars and lollipops that he can buy is given by:

$0.45x + 0.25y \leq 28$

<h3>What is the inequality that models this situation?</h3>

The total price can be no more than $28, hence:

$T \leq 28$

Each candy bar costs $0.45 and each lollipop costs $0.25. x is the number of candy bars and y of lollipops. Hence, the total price is given by:

T = 0.45x + 0.25y.

Hence, the inequality that models the situation is:

$0.45x + 0.25y \leq 28$

More can be learned about inequalities at brainly.com/question/25235995

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x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco

igomit [66]

Answer:

$x=-cos(t)+2sin(t)$

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

$a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0$

Will have a characteristic equation of the form:

$a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0$

Where solutions $r_1,r_2...,r_n$ are the roots from which the general solution can be found.

For real roots the solution is given by:

$y(t)=c_1e^{r_1t} +c_2e^{r_2t}$

For real repeated roots the solution is given by:

$y(t)=c_1e^{rt} +c_2te^{rt}$

For complex roots the solution is given by:

$y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)$

Where:

$r_1_,_2=\lambda \pm \mu i$

Let's find the solution for $x''+x=0$ using the previous information:

The characteristic equation is:

$r^{2} +1=0$

So, the roots are given by:

$r_1_,_2=0\pm \sqrt{-1} =\pm i$

Therefore, the solution is:

$x(t)=c_1cos(t)+c_2sin(t)$

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of $x(t)$ in order to find the constants $c_1$ and $c_2$ :

$x'(t)=-c_1sin(t)+c_2cos(t)$

Evaluating the initial conditions:

$x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1$

And

$x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2$

Now we have found the value of the constants, the solution of the second-order IVP is:

$x=-cos(t)+2sin(t)$

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3 years ago

What two numbers multiplies to 36 and add to negative 21

anastassius [24]

Factor 36 and add them, make sure both are negatie
1+36=37, nope
2+18=20 nope
3+12=15,nope
4+9=13, nope
6+6=12, nope

the closest one is 20

I hope you meant adds to -20, because if you did, then answer is -2 and -18

if you did mean 36 and adds to -21, then the numbers are $\frac{-21-3 \sqrt{33} }{2}$ and $\frac{-21+3 \sqrt{33} }{2}$ , aprox -19.1168 and -1.8831

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