CPhill

The quadrants are correct but the angles should be 161.565051177078° and 341.565051177078° since the angles lie in those quadrants.

On the on-site calculator, hit the "2nd" key and the "atan" button. Make sure you specify degrees or radians (found at the bottom left). "Atan" stands for "arctangent," which is the same thing as the "tangent inverse.'

He will pay a $70 administrative fee for both items and a $35 delivery charge for both items = $105.  Since you didn't specify a tax rate, we can't calculate the taxes due. All we can say definitively is that he will pay at least $105 at the time of purchase.

$${\mathtt{1}}{\mathtt{\,-\,}}{\left({\frac{{{\mathtt{7}}}^{{\mathtt{1}}}}{{\mathtt{10}}}}\right)}^{{\mathtt{50}}} = {\mathtt{0.999\: \!999\: \!982\: \!015\: \!349\: \!6}}$$

Or did you mean this???

$${\mathtt{1}}{\mathtt{\,-\,}}{\left({{\mathtt{7}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{10}}}}\right)}\right)}^{{\mathtt{50}}} = -{\mathtt{16\,806.000\: \!000\: \!000\: \!000\: \!000\: \!9}}$$

6t² = 3    Divide both sides by 6

t² = (1/2)      Take the square root of both sides

t = ±√(1/2)    (If you want to "rationalize" the denominator, we get  ±[√(2)] / 2 .

Remember we have to take both roots when applying the square root property in solving equations !!

OK...let's see if it's "factorable.'

(3x +1) (x + 5)    Nope !!

(3x +5) (x +1)     Nope !!

These are the only possibilities we have, because 3 and 5 are prime.

Thus, the expression isn't "factorable."

A famous question in Zen practice........it's the dichotomy of "two and one". We understand "duality." What then, is "non-duality??"

I'll answer your question if you will answer mine......

In a certain village, every man who doesn't shave himself is shaved by the barber.

The question is.....Who shaves the barber??

For, if he doesn't shave himself, then he is shaved by himself. And, if he does shave himself, then he is not shaved by himself.......

I've thought about this problem for several days. Here's my take on the answer.

As Rom pointed out, the interval [-(7/2), (1/4)] does indeed make the "log" in the numerator undefined.

But I believe the answer to the problem lies in the denominator. In effect, we have the domain of an "inside" function -  (4x-1) - to consider, and the domain of an "outside" function - log(4x-1) - to consider. Note that any values between (1/4) and (1/2) are OK - they just make the denominator negative. But also notice that any values less than (1/4) aren't defined for log(4x-1). And x=(1/2) isn't defined, either.

So, I believe the correct answer to this problem are the two intervals (1/4, 1/2) and (1/2, ∞)

Anybody have another take??

Your (linear) equation has the form y = mx + b, where "m" is the slope and "b" is the y intercept.

So m = -2

And that's the slope.

Without knowing what values  x, y and z  take on, there's nothing to "evaluate."