All Answers 358149 Answers

Minimum occurs when b + 5 = 0, or b = -5

.

.

a = 3,  b = 2  There aren't many possibilities to test!

.

Thank you Anonymous!

HAHAHAHAHAHA! that is hilarious. Nice one rosala

Yes I should have thought of that 

Anon you are doing calculus - surely you can put brackets in properly !!

$$\\For\;\; f(x)\qquad x\ne4\;\;and\;\;y\ne1\\\\
$so for $f^{-1}(x)\qquad
x\ne1 \;\;and\;\;y\ne4\\\\
$that is,$ \\$$

$$f^{-1}(x)$$       is undefined when x=1

Thanks alan!!

f(g(x)) = f(Bx + A) → A(Bx + A) + B → ABx + A² + B

g(f(x)) = g(Ax + B) → B(Ax + B) + A → BAx + B² + A

f(g(x)) - g(f(x)) = A² - B² + B - A → (A - B)(A + B) - (A - B) → (A - B)(A + B -1) 

so we must have (A - B)(A + B -1) = B - A

Since B ≠ A we can divide both sides by A - B to get A + B - 1 = -1, so A + B = 0

.

I think this must be:

$$\frac{d}{dx}(\frac{x}{1+x^2})=\frac{1-x^2}{(1+x^2)^2}$$

which is correct, but the original needs brackets to make it so:

d/dx(x/(1+x²)) = (1-x²)/(1+x²)²

I do not know what you are asking

but

$$\\\frac{d}{dx}(\frac{x}{1}+x^2)\\\\
=\frac{d}{dx}(x+x^2)\\\\
=1+2x\\\\$$

what is the other half for?

Thanks Alan

Here's your free-body diagram triple-xv:

Resolve forces vertically:  Ncosθ = mg

Resolve forces horizontally: mv²/r = Nsinθ

Solve for r.

.

Your Welcome Melody!

Welcome to the forum Mataxa   

I have given you the final answer that you need but do you need more working.

I did almost all of it with online calculators. 

Thanks Rosala. :)

Hookes law says force is proportional to extension. F = k*x

1)  Extension = 10cm (= 0.1m), force = 43N, so k = 43/0.1 N/m  or k = 430N/m

2) Work done is the integral of force *distance with respect to distance

 $$work=\int_0^Xkxdx=\frac{1}{2}kX^2$$

work = (1/2)*430*0.1²J or work = 21500 J

Most of the others can be answered using a similar approach with Hookes law.  The exception is the last one:

10) Force = mass*acceleration

464 = 63*a

a = 464/63 m/s² 

a ≈ 7.365 m/s²

.

yes Melody ive looked a them all!and like if im a bit stressed out or tired then after reading the wrap , it just cheers me up and they are so imformative and tell me what all happened in the day so thats why i love them all!

Like this only one... But there is other 9 hardcore problems...

Slightly simpler way of doing (a) is to divide both sides by cos(p+15°) to get tan(p+15°) = 3

p+15° = tan^-1(3)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{3}}\right)} = {\mathtt{71.565\: \!051\: \!177\: \!078^{\circ}}}$$

so p = 71.565° - 15° 

or p = 56.6° (to one decimal place)

That's the first quadrant solution.  Add 180° to get the third quadrant solution.

.

Thank you Rosala.  That makes me very happy.

Why do you like them?   Do you actually look at all the posts?

How much of this do you need to do by hand?

Well here is a relevant graph

so this is as far as I got... And btw the question is Find all pairs of real numbers I don't know if i translate that good because English isn't my native language...

By using the right formula and not getting so panicked about it!