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3 years ago

Solve the following word problem by using a system of equations.

1 answer:

4 0

For this problem, we can use systems of equations. I will use the variables x (for ice-cream) and y (for soda). We get the system:

2.25x+0.75y=30.00
x+y=18

Putting the second equation in terms of y, we get that y=-x+18. We can substitute this into our first equation.

2.25x+0.75y=30.00
becomes
2.25x+0.75(-x+18)=30

Solving for x, we get that x=11. This satisfies answer B.

However, if you want to check to see if this is correct, you could find y by plugging in your value of x into the first equation, then check to see if your found values satisfy BOTH equations.

2.25(11)+0.75y=30
24.75+0.75y=30
0.75y=5.25
y=7

Then we plug this into both equations to see if they are true.

2.25(11)+0.75(7)=30
24.75+5.25=30
30=30 (This is true)

x+y=18
11+7=18
18=18 (This is true).

Both equations are true, so the value for x and y are correct. We see that only answer B is supported through analyzing your work.

:)

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Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

4 years ago

Which expression is equivalent to xy^((2)/(9))?

bonufazy [111]

Pretty sure it's D. When you have a fractional exponent, the root number is always on the bottom, while the exponent on the top. Since x and y are not grouped together in a parentheses, we can assume that only "y" is being raised to the 2/9 power. That means "x "will stay out in front and y will be in the 9th root being raised to the 2nd power.

7 0

3 years ago

Read 2 more answers

What is the answer? Will give brainilist.

Elina [12.6K]

Answer:

Step-by-step explanation:

first put them in order

5,6,6,8,9,9

now the ones that are in the middle are 6 and 8 so we find the average by adding them then dividing by 2

6+8=14/2=7

5 0

3 years ago

NO LINKS!! Which type of comic does not have vertices or foci?

creativ13 [48]

Answer:

b) Circle

Step-by-step explanation:

All of the conic sections have vertices and foci. These features are not usually talked about for a circle, so perhaps "circle" is the expected answer.

A circle is a special case of ellipse with eccentricity 0. Its foci are coincident at its center, and its vertices are the ends of any pair of perpendicular diameters.

4 0

2 years ago

Which expressions are equivalent to the expression 4a – 6b + 3c? a + 3(a − 2b + 3c) 4a + 3(2b + c) 2(2a − 3b + c) + c 2(2a − 3b)

Dvinal [7]

It is D because (2)(2)=4 and (2)(3)= 6

6 0

4 years ago

Read 2 more answers