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3 years ago

Using the Distance Formula Find the length given the points (0,0) and (0,100)

Mathematics

1 answer:

Ahat [919]3 years ago

7 0

The length is 10
i showed my steps on the picture attached
hope this helps :)

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Which binomial is a difference of squares?

GenaCL600 [577]

D) 36x^2-81
both 36x^2 and -81 are perfect squares
difference means subtraction

7 0

3 years ago

I know how to set it up but just don’t remember how to solve it. V stands for volume.

Tasya [4]

Answer:

29.5788

Step-by-step explanation:

Multiply all of the terms together

5 0

4 years ago

Read 2 more answers

Anyone help please I need it

andreev551 [17]

Answer:

$-10+i\sqrt{2}$

Step-by-step explanation:

One is given the following expression:

$\sqrt{4}-\sqrt{-98}-\sqrt{144}+\sqrt{-128}$

In order to simplify and solve this problem, one must keep the following points in mind: the square root function ( $\sqrt{}$ ) is a way of requesting one to find what number times itself equals the number underneath the radical sign. One must also remember the function of taking the square root of a negative number. Remember the following property: ( $\sqrt{-1}=i$ ). Simplify the given equation. Factor each of the terms and rewrite the equation. Use the square root property to simplify the radicals and perform operations between them.

$\sqrt{4}-\sqrt{-98}-\sqrt{144}+\sqrt{-128}$

$\sqrt{2*2}-\sqrt{-1*2*7*7}-\sqrt{12*12}+\sqrt{-1*2*8*8}$

Take factors from out of under the radical:

$\sqrt{2*2}-\sqrt{-1*2*7*7}-\sqrt{12*12}+\sqrt{-1*2*8*8}$

$2-7i\sqrt{2}-12+8i\sqrt{2}$

Simplify,

$-10+i\sqrt{2}$

8 0

3 years ago

A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro

pshichka [43]

Answer:

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed $\frac{dx}{dt}$ = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

$(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=(x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=rx$

$\frac{dh}{dt}=\frac{rx}{h}$

When automobile is 30 miles past the intersection,

For x = 30

$\frac{dh}{dt}=\frac{30r}{h}$

Since $h=\sqrt{d^{2}+(30)^{2}}$

Therefore,

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}$

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

3 0

3 years ago

Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions

monitta

I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take $z=y'$ , so that $z'=y''$ and we're left with the ODE linear in $z$ :

$y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2$

Now suppose $y$ has a power series expansion

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Then the ODE can be written as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0$

$\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0$

All the coefficients of the series vanish, and setting $x=0$ in the power series forms for $y$ and $y'$ tell us that $y(0)=a_0$ and $y'(0)=a_1$ , so we get the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}$

We can solve explicitly for $a_n$ quite easily:

$a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}$

and so on. Continuing in this way we end up with

$a_n=\dfrac{a_1}{n!}$

so that the solution to the ODE is

$y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x$

We also require the solution to satisfy $y(0)=a_0$ , which we can do easily by adding and subtracting a constant as needed:

$y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

4 0

3 years ago