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The formula for the radius of an inscribed circle in a triangle is 2 * Area= Perimeter * Radius. In an 8, 15, 17 right triangle, twice the area is 8 * 15= 120 and the perimeter is 8+15+17= 40. So the radius is 120/40=3. The radius is half the diameter so your answer is 3 * 2= 6.

No just at that one point. At the point of inflextion the gradient is neither increasing nor decreasing. 

If ithe rate of change is not + or - then it must be 0. Which means that  the gradient must be a constant at that point. 

It is not something I would usually add.  (I can see the confusion)

I would normally have just left the statement out altogether.

It is true though.

This question keeps coming up!  See my solution at https://web2.0calc.com/questions/help-please_7362#r4 

(No need for an approximate numerical method).

a constant where? on the entire line?

If the gradient is not increasing and not decreasing then it must be a constant. 

At a point of inflection

\(\frac{d^2y}{dx^2}=0\\~ \\and\\~\\ \frac{dy}{dx}=\int \frac{d^2y}{dx^2}dx=\int 0dx=k \)

I am not sure that I have presented this properly.

Feel free to comment Max.

what do you mean by f' is constant when f has a point of inflexion?

\(\dfrac{12\sqrt 2 \operatorname{cis}\left(7\pi/6\right)}{3 \sqrt{\operatorname{cis}\left(\pi/4\right)}} = \dfrac{12\sqrt 2 \operatorname{cis}\left(7\pi/6\right)}{3 \operatorname{cis}(\pi/8)} = 4 \sqrt 2 \operatorname{cis}\left(7\pi/6 - \pi/8\right) = 4\sqrt{2} \operatorname{cis} (25\pi/24) = -4\sqrt2 \operatorname{cis}(\pi/24)\)

\(\cos \angle YZX = \dfrac{5^2 + 8^2 - 7^2}{2(5)(8)} = \dfrac12\)

Therefore \(\angle YZX = 60^\circ\).

Area = \(\dfrac12 \cdot 5 \cdot 8 \sin 60^\circ = 10 \sqrt 3\)

By tangent properties,

IL = IJ --- (1)

Also LO = LO --- (2) (obvious)

Also IO and JO are both radii, so IO = JO --- (3)

By (1), (2), and (3), and using SSS postulate, \(\triangle LIO \cong \triangle LJO\).

Therefore \(\angle ILO = \angle JLO\)

\(\implies \) LO bisects ∠ILJ

In spherical coordinates, the \(\phi\) represents the angle between (the line segment between the origin and a point) and the x-y plane.

If \(\phi = c\), that means the angle \(\phi\) is constant. Construct a line L such that the angle between L and the x-y plane is \(\phi\), then revolve it around the z-axis. For any line segment L₀ in the resulting figure, the angle between L₀ and x-y plane is \(\phi\). Therefore the resulting figure is a cone.

Therefore the answer is (F).

\(2x^{2001} + 3x^{2000} + 2x^{1999} + 3x^{1998} + \cdots + 2x + 3 = 0\\ (2x + 3)(x^{2020} + x^{2018} + \cdots + 1) = 0\)

Because \(x^{2020} + x^{2018} + \cdots + 1 > 0\), the only possibility is \(2x + 3 = 0\).

Solving gives \(x = -\dfrac32\).
There is only 1 real root.

Assuming \(\sin^{-1} x \) and \(\tan^{-1} x \) means inverse trigonometric functions,

\(\sin\left(\sin^{-1} \dfrac35 + \tan^{-1} 2\right)\\ = \sin\left(\tan^{-1} \dfrac34 + \tan^{-1}2 \right)\\ = \sin \tan^{-1} \dfrac34 \cos \tan^{-1} 2 + \cos \tan^{-1} \dfrac34 \sin \tan^{-1} 2\\ = \dfrac35 \dfrac1{\sqrt5} + \dfrac45\dfrac{2}{\sqrt5}\\ = \dfrac{11}{5\sqrt5}\)

\(\log_{y^5}\left(x^3\right) = \dfrac{\log x^3}{\log y^5} = \dfrac{3 \log x}{5 \log y} = \dfrac35 \cdot \log_{y} x\)

Therefore k = 3/5.

Lateral area = curved surface area = pi*(8/2)*sqrt(4^2 + 5^2) ~= 80.5 ft^2

\(m \angle KML = m \stackrel{\mbox{$\frown$}}{KL} \;= 98^\circ\)

Because the angle at center is twice the angle at circumference,

\(m\angle KJL = \dfrac12 m\angle KML = 49^\circ\)

c + di lies on the 1st Quadrant, which means the polar form of c + di is \(re^{i\theta}\), where \(r = \sqrt{c^2 + d^2}\), \(\theta \in \left(0, \dfrac\pi2\right) \).

e - fi lies on the 4th Quadrant, which means the polar form of e - fi is \(Re^{i\phi}\), where \(R = \sqrt{e^2 + f^2}\), \(\phi \in \left(\dfrac{3\pi}2, 2\pi\right) \)

Therefore, (c + di)(e - fi) = \(rRe^{i\left(\theta + \phi\right)}\)

For \(\theta \in \left(0, \dfrac\pi2\right) \) and \(\phi \in \left(\dfrac{3\pi}2, 2\pi\right) \), \(\theta + \phi \in \left(\dfrac{3\pi}{2} , \dfrac{5\pi}{2}\right)\)

Therefore (c + di)(e - fi) can lie inside the 4th Quadrant and the 1st Quadrant.

a)

\(f'(x) \text{ is }\left\{\begin{matrix}\text{positive}\\\text{negative}\end{matrix}\right\}\text{ when }f(x)\text{ is }\left\{\begin{matrix}\text{strictly increasing}\\\text{strictly decreasing}\end{matrix}\right\}.\)

\(\implies f(x) \text{ is strictly increasing on the interval }(-\infty, 1)\cup (3, \infty)\text{ and strictly decreasing on the interval } (1, 3).\)

What about when x = 1 or x = 3?

Let ε be an arbitrary constant, infinitesmally small.

\(f'(1 -\varepsilon) > 0\text{ and }f'(1 + \varepsilon) < 0\)

At x = 1, local maximum of f(x) occurs.

\(f'(3 -\varepsilon) < 0\text{ and }f'(3 + \varepsilon) > 0\)

At x = 3, local minimum of f(x) occurs.

\(f'(x) \text{ is }\left\{\begin{matrix}\text{increasing}\\\text{decreasing}\\\text{constant}\end{matrix}\right\}\text{ when }f(x)\text{ }\left\{\begin{matrix}\text{is convex}\\\text{is concave}\\\text{has a point of inflexion}\end{matrix}\right\}.\)

\(\implies f(x) \text{ is concave on the interval } (-\infty, 2)\text{ and is convex on the interval }(2, \infty).\)

\(\text{Point of inflexion of }f(x)\text{ occurs at } x =2.\)

b)

\(f'(x) \text{ is }\left\{\begin{matrix}\text{increasing}\\\text{decreasing}\\\text{constant}\end{matrix}\right\}\text{ when }f''(x)\text{ is }\left\{\begin{matrix}\text{positive}\\\text{negative}\\\text{0}\end{matrix}\right\}.\)

\(\implies f''(x) \text{ is negative on the interval } (-\infty, 2)\text{ and is positive on the interval }(2, \infty).\)

\(f''(2) = 0\)

You have two sides and the included angle, so you can find the area by using \(\text{Area} = \dfrac12 bc \sin \theta\), where b and c are the side lengths and \(\theta\) is the included angle.

What is the approximate measure of the angle formed by this new guy wire and the ground?

This equation cannot be solved with algebra, and you can only approximate its roots by Newton-Raphson iteration.

If \(\mathbf A^2 = \mathbf A\), then \(\left(\det \mathbf A\right)^2 = \det \mathbf A\),

which means \(\det \mathbf A\) = 0 or 1

Let f(x) = x(x - 1)(x - 2) Q(x) + R(x), where Q and R are polynomials which represents quotient and remainder respectively.

Let R(x) = ax^2 + bx + c.

f(0) = 4

R(0) = 4

c = 4

f(1) = 5 

R(1) = 5

a + b + c = 5

a + b = 1 --- (1)

f(2) = 10

R(2) = 10

4a + 2b + c = 10

4a + 2b = 6 --- (2)

(2) - 2 * (1) : 2a = 4

a = 2

2 + b = 1

b = -1

Therefore

R(x) = 2x^2 - x + 4

arc EH = arc FG

\(x^2 - 2x = 56 - 3x\\ x^2 + x - 56 = 0\\ (x + 8)(x - 7) = 0\\ x = -8\text{(rej.) or }x = 7\\ \)

Both arcs are 35 degrees.

arc EPF = \(360^\circ - 70^\circ - 2\cdot 35^\circ = \boxed{220^\circ}\)

By power of points,

\(4(4 + 18) = 6(6 + x)\\ x = \dfrac{26}3\)

By Lagrange multipliers,

\(\begin{cases} \nabla_{x,y,z}{\left(4x^2 + y^2 + 16z^2 - 1\right)} = \lambda \nabla_{x,y,z}\left(7x + 2y + 8z\right)\\ 4x^2 + y^2 + 16z^2 = 1 \end{cases}\)

\(\begin{cases} x = \dfrac78 \lambda\\ y = \lambda\\ z = \dfrac14\lambda\\ 4x^2 + y^2 + 16z^2 = 1 \end{cases}\)

Substitute the first three equations to the fourth one,

\(\dfrac{49}{16}\lambda^2 + \lambda^2 + \lambda^2 = 1\\ \lambda^2 = \dfrac{16}{81}\\ \lambda = \pm \dfrac 49\)

Optimum occurs when \((x,y,z) = \left(\dfrac7{18}, \dfrac49, \dfrac19\right) \) or \((x,y,z) = \left(-\dfrac7{18}, -\dfrac49, -\dfrac19\right) \)

Substitute these values to find that the maximum value of 7x + 2y + 8z occurs when \((x,y,z) = \left(\dfrac7{18}, \dfrac49, \dfrac19\right) \), and the value is \(\dfrac92\).