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Rudik [331]
3 years ago
5

What is the slope of the line that contains the points (-1,2) and (2,2)

Mathematics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

the equation to find the slope is (y2-y1)/(x2-x1)

the points are (-1,2) and (2,2) so plug it in and you get

(2-2)/(2-(-1))

0/ (2+1)

0/3

whenever the numerator is 0, that means the slope is 0. the line is just a horizontal line.

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What is the pattern in the values as the exponents increase?
nirvana33 [79]

Answer:

"Multiply the previous value by 2"

Step-by-step explanation:

Let's check the first 2 terms for all the answer choices.

1. Add \frac{1}{16} to the previous value:

\frac{1}{32}+\frac{1}{16}=\frac{3}{32}

Doesn't match.

2. Subtract \frac{1}{16} from the previous value:

\frac{1}{32}-\frac{1}{16}=-\frac{1}{32}

Doesn't match.

3. Divide the previous value by 2:

\frac{\frac{1}{32}}{2}\\=\frac{1}{32}*\frac{1}{2}=\frac{1}{64}

Doesn't match.

4. Multiply the previous value by 2:

\frac{1}{32}*2=\frac{2}{32}=\frac{1}{16}

DOES WORK!

Also, doing the same thing with all the other values would give us matching answer. So this choice is right.

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3 years ago
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Luda [366]

Answer:

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Step-by-step explanation:

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3 years ago
Witch multiples of 24 are not factors of 24
Nataly [62]

Answer:

A factor of 24 is a number that can be multilied by another number (normally whole (not fraction or decimal) numbers) so a factor must be less than the number any multiple of 2 greater than 24 will work so just any even number greater than 24 exg 26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58,60, and so on.

Step-by-step explanation:


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Answer:

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Step-by-step explanation:

7 0
3 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
4 years ago
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