Recent content by mrtwhs

We are given that $x>y$ and $w>v$ therefore $x+w > y+v = \angle ABC$ Now $\angle CAB > x$ and $\angle ACB >w$ so $\angle CAB + \angle ACB > x+w > y+v = \angle ABC$ But $\angle CAB + \angle ACB = 180 - \angle ABC$ so $180 - \angle ABC > \angle ABC$ and $\angle ABC < 90$

$u=x-3$

If $y=x+\sin x$, then the reflection around $y=-x$ is $-x=-y+\sin(-y)$ or $x=y+\sin y$. Although a function, this cannot be explicitly solved for $y$. However the two functions are also symmetric around $y=x$ and intersect at the point $(2\pi,2\pi)$. So the area under $g$ between your limits...

Although $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$ is a line, $3\beta^2x+3(1-2\alpha)y+z=3$ is not. It is a plane.

In your diagram, all three sides of $\triangle AEC$ are known. Using the law of cosines you get $2\alpha = \cos^{-1}(7/8)$ and $\alpha \approx 14.47751219^{\circ}$. Continuing to use the law of cosines we get $AD=\dfrac{3\sqrt{10}}{2}$ and $AB=3\sqrt{6}$. Finally, using the law of cosines on...

Rewrite the first equation as $y=\dfrac{M-x^2}{2x}$ and then substitute into the second equation. After quite a bit of multiplying, you will get $5x^4-(2M+4N)x^2+m^2=0$. Use the quadratic formula to solve for $x^2$ and you will get the value for $x^2$ given in the statement. The value for...

Letting $AB=c$, $AC=b$ and noting that the usual property of the centroid tells us that $GC = \frac{2}{3}$ of the median from $C$ and $GB = \frac{2}{3}$ of the median from $B$ we can use Stewart's Theorem to write everything in terms of the three sides $a,b,c$. It is an involved equation...

Why does it have to be 16 or 17? Just call it $$16\dfrac{2}{3}$$. There is no rule which says that expected value must be an attainable value (unless it is expressly stated at the start of the problem). Expected value is essentially the long term average. The analogy would be a math class...

HallsofIvy has given the correct advice for solving this equation ... but ... I'm going to take an epsilon amount of issue with one phrase ... "anything we do to one side of the equation we must do to the other". I've said this myself in Algebra classes but I paid for it once. I had a student...

Send me to the woodshed! I misplaced my parentheses.

$$\ln(m-n) = \ln \left( \dfrac{m}{n} \right) \neq \dfrac{\ln m}{\ln n}$$

Wouldn't vectors 1,3,4,5 be an equally good answer?

Double check your arithmetic. With $$u=4$$, $$v=15$$ and $$a=24$$, you should get the equation $$w^4-482w^2+52897=0$$ which factors into $$(w-13)(w+13)(w^2-313)=0$$ This appears to yield two possible answers $$w=13$$ and $$w=\sqrt{313}$$.

$$k = \dfrac{u+v+w}{2}$$, $$k-u = \dfrac{-u+v+w}{2}$$, $$k-v = \dfrac{u-v+w}{2}$$, $$k-w = \dfrac{u+v-w}{2}$$. Substituting in we get $$a^2=\dfrac{1}{16}(u+v+w)(u+v-w)(w+(u-v))(w-(u-v))$$ Simplifying yields $$16a^2=((u+v)^2-w^2)(w^2-(u-v)^2)$$ $$-16a^2=(w^2-(u+v)^2)(w^2-(u-v)^2)$$...

When you say $$x=3$$, that means you put the 3 where the $$x$$ is! $$3=1.5y$$. Now try it.