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3 years ago

Is it rise over run or run over rise I get it's most likely easy but I don't trust my brain.

Mathematics

2 answers:

sammy [17]3 years ago

7 0

rise over run

you can remember it doing the vertical over horizontal

MrMuchimi3 years ago

7 0

Answer:

Step-by-step explanation:

It’s rise over run

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The box plots show the weights, in pounds, of the dogs in two different animal shelters.

Dmitriy789 [7]

Answer:

The median weight for shelter A is greater than that for shelter B.

The data for shelter B are a symmetric data set.

The interquartile range of shelter A is greater than the interquartile range of shelter B.

Step-by-step explanation:

The median weight for shelter A is greater than that for shelter B.

The median of A = 21 and the median of B = 18 true

The median weight for shelter B is greater than that for shelter A.

The median of A = 21 and the median of B = 18 false

The data for shelter A are a symmetric data set.

False, looking at the box it is not symmetric

The data for shelter B are a symmetric data set.

true, looking at the box it is symmetric

The interquartile range of shelter A is greater than the interquartile range of shelter B.

IQR = 28 - 17 = 11 for A

IQR for B = 20 -16 = 4 True

7 0

3 years ago

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30% of number A is the same as 40% of number B. what could A and B ? <br><br>

Sunny_sXe [5.5K]

9514 1404 393

Answer:

A = 4, B = 3

Step-by-step explanation:

You require that ...

0.30 × A = 0.40 × B

Then ...

B = (3/4)A . . . . . divide by 0.40

Any pair of numbers that satisfies this equation will be possible values of A and B. One such pair is (A, B) = (4, 3).

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2 years ago

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What is the measure of angle BAC? Round to the nearest

tigry1 [53]

Answer:

the answer is c. 44, i hope it works

Step-by-step explanation:

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3 years ago

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What is 9/10 - 1/3 Closer to 0,1,1/2

AveGali [126]

Answer:

1/2

Step-by-step explanation:

9/10-1/3=0.56666666666666666...

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3 years ago

Read 2 more answers

Help with q25 please. Thanks.

Westkost [7]

First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

$f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\$

Now apply the derivative to that to get the second derivative

$f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\$

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

-----------------------------------

Let's compute dy/dx. We'll use f(x) as defined earlier.

$y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\$

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\$

If you need a refresher on the quotient rule, then

$\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\$

where P and Q are functions of x.

-----------------------------------

This then means

$\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\$

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

------------------------------------

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form ( f(x) )^2 in the numerator to cancel out with the denominator.

$\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\$

So this concludes the proof that $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\$ when $y = \ln\left(\sin(px)+\cos(px)\right)\\\\$

Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

7 0

2 years ago