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4 years ago

Given H(t)=10-2t^2 , find h^-1 (t)

Mathematics

1 answer:

kobusy [5.1K]4 years ago

5 0

I’m
Not really sure to this question but I think 2+

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Here’s Question three

Hitman42 [59]

Answer:

BD = 3 units

Step-by-step explanation:

Since, AD is an angle bisector of ∠BAC,

m∠BAD = m∠CAD = 20°

CD = 3 units

In ΔACD and ΔABD,

m∠BAD = m∠CAD = 20° (Given)

AD ≅ AD [Reflexive property]

Therefore, by H-A property of congruence both the triangles will be congruent.

And by CPCTC,

CD ≅ BD = 3 units

5 0

3 years ago

22= 18x + 4 please help!!

vredina [299]

Answer:

x=1

Step-by-step explanation:

22-18=4 18-18=/

18x1=18

22=18(1)+4

22-18(1)=4

7 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B12%7D%7Bx%7D%20%20%5C%5C%20y%20%3D%20%20%5Cfrac%7B6%7D%7B%20%7Bx%7D%5E%

soldi70 [24.7K]

Answer:

x = 1/2

y = 24

Step-by-step explanation:

y = 12 / x

y = 6 / x²

12 / x = 6 / x²

x² / x = 6 / 12

x = 1 / 2

y = 12 / x

y = 12 / (1/2)

y = 24

4 0

3 years ago

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

3 years ago

Convert 6 1/4 lb to ounces.

poizon [28]

Answer:

B 100 ounces

Step-by-step explanation:

There's 16 ounces in a pound. So multiply 6.25 by 16 to get 100.

7 0

2 years ago