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sammy [17]
2 years ago
10

Can someone please help me with this

Mathematics
1 answer:
julsineya [31]2 years ago
8 0

Answer:

Step-by-step explanation:

5- There are 2 right-angled triangles and 1 equilateral triangle or you can just say triangles

6- area for a triangle is a=bh(1/2)

so 4×3=12(1/2)=6

and 4×3=12(1/2)=6

In both triangles, they're a=6

7- 4×6=24(1/2)=12

the top flag is a=12

8-12+6+6=24

9- the rectangular flag has an area of 24 since a=bh and the added areas of the triangles are 24. I conclude that these are two ways of finding the area of this rectangular flag.

hope this helps : )

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If the given triangles FGH and TUV are congruent, their corresponding sides will be equal in measure.

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3 years ago
Place labels on the coordinate plane below.
Nuetrik [128]

Answer:

the x-axis is the middle horizontal line

the y-axis in the middle vertical line

the origin is (0,0) or where the x and y-axis cross

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in the upper right quadrant the x-axis is positive and so is the y.

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8 0
3 years ago
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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
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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.

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\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)

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

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

We then end up with

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