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2 years ago

12

Plz help it’s due tonight!!!

1 answer:

gogolik [260]2 years ago

3 0

Answer:

x = 12

A = 83

B = 38

c = 59

Step-by-step explanation:

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You are in a bike race. When you get to the first checkpoint you are 2/5 of the distance away to the second checkpoint. When you

gizmo_the_mogwai [7]

the answer is 1/4 of a distance away

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2 years ago

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$24 saved after 3 weeks; $52 saved after 7 weeks. are these ratios equivalent

SIZIF [17.4K]

To figure out this question, I first divided 24 into 3, which gives me an answer of 8. then I divided 52 into 7 , which also is 8 . because the two numbers are the same, I know the ratios are equivalent.

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3 years ago

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An isosceles triangle’s altitude will bisect its base. Which expression could be used to find the area of the isosceles triangle

omeli [17]

Answer:

$\dfrac{\sqrt{40}\cdot \sqrt{40}}{2}$

Step-by-step explanation:

The length of the base is the distance between the points 4+2i and 10+4i, so

$\text{Base}=|10+4i-(4+2i)|=|10+4i-4-2i|=|6+2i|=\sqrt{6^2+2^2}=\\ \\=\sqrt{36+4}=\sqrt{40}$

The middle point of the base is placed at point

$\dfrac{4+2i+10+4i}{2}=\dfrac{6i+14}{2}=7+3i$

The length of the height is the distance between the points 5+9i and 7+3i

$\text{Height}=|5+9i-(7+3i)|=|5+9i-7-3i|=|-2+6i|=\sqrt{(-2)^2+6^2}=\\ \\=\sqrt{4+36}=\sqrt{40}$

So, the area of the triangle is

$A=\dfrac{1}{2}\cdot \text{Base}\cdot \text{Height}=\dfrac{\sqrt{40}\cdot \sqrt{40}}{2}$

4 0

2 years ago

Round the numbers to the nearest tenth, then find the sum:<br> 3.296 + 0.9785.

Debora [2.8K]

Answer:

4.3

Step-by-step explanation:

first round

3.296 is 3.3

0.9785 is 1

3.3 + 1 = 4.3

4 0

2 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

3 years ago