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4 years ago

Given a vector with the initial point, (4,2) and the terminal point, (-3,1), find the vector in component form.

Mathematics

1 answer:

ioda4 years ago

7 0

$\bf (-3,1),(4,2)\implies \implies \implies$

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Consider the differential equation,

kondor19780726 [428]

Answer:

The two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

Step-by-step explanation:

As the given equation is

$y''+7y'-8y=0\\$

So the corresponding equation is given as

$m^2+7m-8=0$

Solving this equation yields the value of m as

$(m+8)(m-1)=0\\m=-8, m=1$

Now the equation is given as

$y(t)=C_1e^{m_1t}+C_2e^{m_2t}$

Here m1=-8, m2=1 so

$y(t)=C_1e^{-8t}+C_2e^{t}$

The derivative is given as

$y'(t)=-8C_1e^{-8t}+C_2e^{t}$

Now for the first case y(t=0)=1, y'(t=0)=0

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\1=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\0=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=1\\-8C_1+C_2=0$

Solving the equation yield

$C_1=1/9 \\C_2=8/9$

So the function is given as

$y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$

Now for the second case y(t=0)=0, y'(t=0)=1

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\0=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\1=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=0\\-8C_1+C_2=1$

Solving the equation yield

$C_1=-1/9 \\C_2=1/9$

So the function is given as

$y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

So the two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

7 0

3 years ago

Is this correct? (Image below). If not, what is the right answer?!?!

ahrayia [7]

Answer:

You're correct great job!

8 0

3 years ago

Read 2 more answers

Write an expression for the calculation subtract 10 divided into fifths from 20 diveded in half

vredina [299]

20/2-10/5. You divide 20 by 2 and subtract that from 10 divided by 5

5 0

3 years ago

What is the h?<br> h/6 = 20/24

Mademuasel [1]

Answer:

h = 5

Step-by-step explanation:

$\frac{h}{6} = \frac{20}{24} \\ 24 \times h = 20 \times 6 \\ 24h = 120 \\ h = \frac{120}{24} \\ h = 5$

5 0

3 years ago

61 random samples of city rent prices are selected from a normally distributed population the samples have a mean of $738 and a

Alenkinab [10]

Answer:

(728, 748)

Step-by-step explanation:

You need the Z value which for 95% confidence interval is 1.96. The mean is $738 and the standard deviation of $41.

$X-Zs/\sqrt{n} \\ X+Zs/\sqrt{n}$

Therefore we can calculated the confidence interval

$738-1.96\cdot{41}/\sqrt{61} =738-10.3$

$738±10.3$

The interval is (728,748)

3 0

3 years ago