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3 years ago

ABCD∼EFGH AD=45 in. , EH=75 in. , and AB=30 in. What is EF ?

1 answer:

3 0

EF = 50 in.

In similar figures, corresponding sides are proportional. We use the similarity statement ABCD~EFGH to write a proportion:

$\frac{AB}{EF}=\frac{AD}{EH}
\\
\\ \frac{30}{x}=\frac{45}{75}$

Cross multiply the proportion:
45*x = 75*30
45x = 2250

Divide both sides by 45:
45x/45 = 2250/45
x = 50

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LiRa [457]

$\bf -\cfrac{2}{5}+\cfrac{2}{3}p+\cfrac{1}{3}p=2p-\cfrac{16}{5}\implies \stackrel{\textit{LCD of 15}}{-\cfrac{2}{5}+\cfrac{2p}{3}+\cfrac{p}{3}}=\stackrel{\textit{LCD of 5}}{2p-\cfrac{16}{5}}
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-30+50p+25p=150p-240\implies -30+240=150p-75p
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8 0

3 years ago

What is the distance between -15.5 and -8.7? A: 24.2 B:7.2. C. 6.8 D. -24.2

garri49 [273]

B. 7.2

Step-by-step explanation:

-15.5+7.2=(-8.7) and vice versa.

5 0

3 years ago

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

The length of a rectangle is 2 units shorter than one-fifth of the width, x.

sergey [27]

Answer:

C. $P=\frac{12x}{5} + 4$

Step-by-step explanation:

Width = x

Length = one fifth of width minus 2

Length = $\frac{1}{5}x-2$

Perimeter is given by the formula

$P=2(l+b)$

$P=2(\frac{1}{5}x-2+x)$

$P=2(\frac{x}{5}-2+x)$

$P=2\times \frac{x-2 \times 5+5 \times x}{5}$

$P=2\times \frac{x-10+5x}{5}$

$P=2\times \frac{6x}{5} - 2\times \frac{-10}{5}$

$P=2\times \frac{6x}{5} + 2\times \frac{10}{5}$

$P=\frac{12x}{5} + 4$

5 0

3 years ago

Is it possible to identify The exact values of all the original service times? A. Yes. The data values in each class are equal t

kakasveta [241]

Answer:

Step-by-step explanation:

The answer to your question would be C. No, the data values in each class could take on any value between the class limits, inclusive.

I hope it helps! Have a great day!

Muffin~

8 0

2 years ago

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