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Sergeu [11.5K]
3 years ago
15

Describe the similarities and differences in the solution of 2x -7 =15 and 2x - 7 < 15

Mathematics
1 answer:
KonstantinChe [14]3 years ago
5 0
<span>Both solutions will have values that exist at the "11" value. However, the major difference is due to one being an equation and the other an inequality. The first equation only has one solution. The inequality has infinitely many solutions, and they are all the values that are less than 11 (x<11).</span>
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zhannawk [14.2K]

4.7 as a fraction would be 4 and 7/10 (7 over 10).

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tester [92]
Set up a proportion.  take 3/6 = 20/x
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Timothy's swim practice started at
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Answer:

1 hour and 15 minutes

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6 0
1 year ago
How do we get 45 in this????
ss7ja [257]

Answer:

A. 45h = 1,575

Step-by-step explanation:

Area of trapezoid = 1,575 cm² (given)

We are given the length of both bases, a = 63 cm and b = 27 cm

We need to find the formula that will enable us determine its height (h).

Thus:

Area of trapezoid = ½(a + b)h

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8 0
2 years ago
Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t&lt;3 if 3≤t&lt;5 if 5≤t&lt;[infinity],y(0)=4. y′+5y={0 if 0≤t&lt;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
3 years ago
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